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\noindent {\bf Question}

\noindent {\bf The series scavenger hunt:} for each of the
infinite series given below, do the following:
\begin{itemize}
\item Determine whether the series converges absolutely, converges
conditionally, or diverges;
\item if the series converges, determine its limit, where possible.
\end{itemize}
\begin{enumerate}
\item $\sum_{n=0}^\infty \frac{2^{n-1}}{3^n}$;
\item $\sum_{n=0}^\infty (1.01)^n$;
\item $\sum_{n=1}^\infty (\frac{e}{10})^n$;
\item $\sum_{n=1}^\infty \frac{1}{n^2+n+1}$;
\item $\sum_{n=1}^\infty \frac{1}{n + \sqrt{n}}$;
\item $\sum_{n=1}^\infty \frac{1}{1+3^n}$;
\item $\sum_{n=2}^\infty \frac{10 n^2}{n^3 - 1}$;
\item $\sum_{n=1}^\infty \frac{1}{\sqrt{37n^3 + 3}}$;
\item $\sum_{n=1}^\infty \frac{\sqrt{n}}{n^2+n}$;
\item $\sum_{n=2}^\infty \frac{2}{\ln(n)}$;
\item $\sum_{n=1}^\infty \frac{\sin^2(n)}{n^2+1}$;
\item $\sum_{n=1}^\infty \frac{n+2^n}{n+3^n}$;
\item $\sum_{n=2}^\infty \frac{1}{n^2\ln(n)}$;
\item $\sum_{n=1}^\infty \frac{n^3+1}{n^4+2}$;
\item $\sum_{n=1}^\infty \frac{1}{n + n^{3/2}}$;
\item $\sum_{n=1}^\infty \frac{10 n^2}{n^4+1}$;
\item $\sum_{n=2}^\infty \frac{n^2 -n}{n^4 +2}$;
\item $\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$;
\item $\sum_{n=1}^\infty \frac{1}{3+5^n}$;
\item $\sum_{n=2}^\infty \frac{1}{n-\ln(n)}$;
\item $\sum_{n=1}^\infty \frac{\cos^2(n)}{3^n}$;
\item $\sum_{n=1}^\infty \frac{1}{2^n+3^n}$;
\item $\sum_{n=1}^\infty \frac{1}{n^{(1+\sqrt{n})}}$;
\item $\sum_{n=1}^\infty 1 / (2^n (n+1))$;
\item $\sum_{n=1}^\infty n! / (n^2 e^n)$;
\item $\sum_{n=2}^\infty \sqrt{n} / (3^n \ln(n))$;
\item $\sum_{n=2}^\infty (2n)! / (n!)^3$;
\item $\sum_{n=1}^\infty (1 - (-1)^n) / n^4$;
\item $\sum_{n=1}^\infty (2+\cos(n)) / (n + \ln(n))$;
\item $\sum_{n=3}^\infty 1 / (n \ln(n) \sqrt{\ln(\ln(n))})$;
\item $\sum_{n=1}^\infty n^n / (\pi^n n!)$;
\item $\sum_{n=1}^\infty 2^{n+1} / n^n$;
\item $\sum_{n=1}^\infty (-1)^{n-1} / \sqrt{n}$;
\item $\sum_{n=1}^\infty \cos(\pi n) / ( (n+1) \ln(n+1) )$;
\item $\sum_{n=1}^\infty (-1)^n (n^2 -1) / ( n^2+1)$;
\item $\sum_{n=1}^\infty (-1)^n / (n \pi^n)$;
\item $\sum_{n=1}^\infty (-1)^n (20n^2 -n -1) / (n^3+n^2+33 )$;
\item $\sum_{n=1}^\infty n! / (-100)^n$;
\item $\sum_{n=3}^\infty 1 / (n \ln(n) (\ln(\ln(n)))^2)$;
\item $\sum_{n=1}^\infty (1 + (-1)^n) / \sqrt{n}$;
\item $\sum_{n=1}^\infty e^n \cos^2(n) / (1+\pi^n)$;
\item $\sum_{n=2}^\infty n^4 / n!$;
\item $\sum_{n=1}^\infty (2n)! 6^n / (3n)!$;
\item $\sum_{n=1}^\infty n^{100} 2^n / \sqrt{n!}$;
\item $\sum_{n=3}^\infty (1+n!) / (1+n)!$;
\item $\sum_{n=1}^\infty  2^{2n} (n!)^2 / (2n)!$;
\item $\sum_{n=1}^\infty (-1)^n / ( n^2 + \ln(n) )$;
\item $\sum_{n=1}^\infty (-1)^{2n} / 2^n $;
\item $\sum_{n=1}^\infty (-2)^n / n!$;
\item $\sum_{n=0}^\infty -n / (n^2+1)$;
\item $\sum_{n=1}^\infty 100\cos(n\pi) / (2n+3)$;
\item $\sum_{n=10}^\infty \sin((n+1/2)\pi) / \ln(\ln(n))$;
\item $\sum_{n=1}^\infty (2n)! / ( 2^{2n} (n!)^2)$;
\item $\sum_{n=1}^\infty (n / (n+1) )^{n^2}$;
\item $\sum_{n=1}^\infty 1 / (1+2+\cdots+n)$;
\item $\sum_{n=1}^\infty \ln(n) / (2n^3 - 1)$;
\item $\sum_{n=1}^\infty \sin(n) / n^2$;
\item $\sum_{n=1}^\infty (-1)^n (n-1) / n$;
\item $\sum_{n=1}^\infty (-1)^n 2^{3n} / 7^n$;
\item $\sum_{n=1}^\infty \cos(n) / n^4$;
\item $\sum_{n=1}^\infty (-1)^n 3^n / (n(2^n + 1))$;
\item $\sum_{n=1}^\infty (-1)^{n-1} n / (n^2+1)$;
\item $\sum_{n=2}^\infty (-1)^{n-1} / (n\ln^2(n))$;
\item $\sum_{n=1}^\infty (-1)^{n-1} 2^n / n^2$;
\item $\sum_{n=1}^\infty (-1)^n \sin(\sqrt{n}) / n^{3/2}$;
\item $\sum_{n=1}^\infty n^4 e^{-n^2}$;
\item $\sum_{n=1}^\infty \sin(n\pi /2) / n$;
\item $\sum_{n=2}^\infty 1 / (\ln(n))^8$;
\item $\sum_{n=13}^\infty 1 /( n\ln(n) (\ln(\ln(n)))^p )$, where $p &gt;0$
is an arbitrary positive real number;
\end{enumerate}
\medskip

\noindent {\bf Answer}

\noindent We make implicit use of the fact that convergence and
absolute convergence are the same for series with positive terms.
\begin{enumerate}
\item {\bf converges absolutely:} we could apply the ratio test, but
we do not need to use such heavy machinary.  Instead, we note that
\[ \sum_{n=0}^\infty \frac{2^{n-1}}{3^n} = \frac{1}{2}
\sum_{n=0}^\infty \frac{2^n}{3^n} =\frac{1}{2} \sum_{n=0}^\infty
\left( \frac{2}{3}\right)^n =\frac{1}{2} \frac{1}{(1-2/3)}
=\frac{3}{2}, \] since $\sum_{n=0}^\infty \frac{2^n}{3^n}$ is a
convergent geometric series.

\item {\bf diverges:} this is a geometric series, and since $1.01 &gt;1$,
it is a divergent geometric series.

\item {\bf converges absolutely:} this is a convergent geometric
series, since $\frac{e}{10} &lt;1$, and it converges to
\[ \sum_{n=1}^\infty \left( \frac{e}{10}\right)^n =\sum_{n=0}^\infty
\left( \frac{e}{10}\right)^n -1 =\frac{1}{1 -e/10} -1 =
\frac{10}{10 - e} - \frac{10 -e}{10 -e} =\frac{e}{10 -e}. \]

\item {\bf converges absolutely:} we use the second comparison test:
since $n^2+n+1 &gt;n^2$ for all $n\ge 1$, we have that
$\frac{1}{n^2+n+1} &lt; \frac{1}{n^2}$ for all $n\ge 1$.  Since
$\sum_{n=1}^\infty \frac{1}{n^2}$ converges, we have that
$\sum_{n=1}^\infty \frac{1}{n^2+n+1}$ converges.

\item {\bf diverges:} note that for $n\ge 1$, we have that $n \ge
\sqrt{n}$, and so $n +\sqrt{n}\le 2n$.  Therefore,
$\frac{1}{n+\sqrt{n}} \ge \frac{1}{2n}$ for $n\ge 1$.  Since the
harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, its
multiple $\sum_{n=1}^\infty \frac{1}{2n}$ diverges, and hence by
the first comparison test the series $\sum_{n=1}^\infty 1 / (n +
\sqrt{n})$ diverges.

\item {\bf converges absolutely:} since $1 + 3^n &gt; 3^n$ for all $n\ge
1$, we have that $\frac{1}{1+3^n} &lt; \frac{1}{3^n}$ for all $n\ge
1$. Since $\sum_{n=1}^\infty \frac{1}{3^n} =\sum_{n=1}^\infty
\left( \frac{1}{3}\right)^n$ converges, the second convergence
test yields that $\sum_{n=1}^\infty 1 / (1+3^n)$ converges.

\item {\bf diverges:} we'll use the limit comparison test: for large
values of $n$, it seems that $\frac{10 n^2}{n^3 - 1}$ behaves like
a constant multiple of $\frac{1}{n}$, and in fact
\[ \lim_{n\rightarrow\infty} \frac{10 n^2/(n^3 - 1)}{1/n}
=\lim_{n\rightarrow\infty} \frac{10 n^3}{n^3 -1} =10 =L. \] Since
the limit exists and $0 &lt; L =10 &lt; \infty$, and since
$\sum_{n=1}^\infty \frac{1}{n}$ diverges, the limit comparison
test yields that $\sum_{n=2}^\infty 10 n^2 / (n^3 - 1)$ diverges.

\item {\bf converges absolutely:} again we'll use the limit comparison
test: for large values of $n$, it seems that $1 / \sqrt{37n^3 +
3}$ behaves like $1/n^{3/2}$, and in fact
\[ \lim_{n\rightarrow\infty} \frac{1 / \sqrt{37n^3 + 3}}{1/n^{3/2}}
=\lim_{n\rightarrow\infty} \frac{n^{3/2}}{\sqrt{37n^3 + 3}}
=\lim_{n\rightarrow\infty} \frac{1}{\sqrt{37 + 3/n^3}}
=\frac{1}{\sqrt{37}} =L. \] Since the limit exists and $0 &lt; L
=\frac{1}{\sqrt{37}} &lt; \infty$, and since $\sum_{n=1}^\infty
\frac{1}{n^{3/2}}$ converges, the limit comparison test yields
that $\sum_{n=1}^\infty 1 / \sqrt{37n^3 + 3}$ converges.

\item {\bf converges absolutely:} we start this one with a bit of
algebra, namely
\[ \frac{\sqrt{n}}{n^2+n} &lt; \frac{\sqrt{n}}{n^2}
=\frac{1}{n^{3/2}}. \] From note 1., we know that
$\sum_{n=1}^\infty 1/n^{3/2}$ converges, and so by the second
comparison test, $\sum_{n=1}^\infty \sqrt{n} / (n^2+n)$ converges.

\item {\bf diverges:} since $\ln(n) &lt;n$ for all $n\ge 2$, we have that
$\frac{1}{\ln(n)} &gt; \frac{1}{n}$ for all $n\ge 2$, and so
$\sum_{n=2}^\infty 2 / \ln(n)$ diverges by the first comparison
test, comparing it to the harmonic series $\sum_{n=1}^\infty
\frac{1}{n}$.

\item {\bf converges absolutely:} since $0 &lt; \sin^2(n)\le 1$ for all
$n\ge 1$, we have that
\[ 0 &lt; \frac{\sin^2(n)}{n^2+1} \le \frac{1}{n^2+1} &lt; \frac{1}{n^2}
\]
for all $n\ge 1$.  Since we are dealing with a series with
positive terms and since $\sum_{n=1}^\infty \frac{1}{n^2}$
converges by note 1., we have that $\sum_{n=1}^\infty \sin^2(n) /
(n^2+1)$ converges by the second comparison test.

\item {\bf converges absolutely:} for this series, we start with a bit
of algebraic massage:
\[ \frac{n+2^n}{n+3^n} &lt; \frac{n+ 2^n}{3^n} &lt; \frac{2^n + 2^n}{3^n} =
2\left( \frac{2}{3}\right)^n. \] So, the second comparison test,
comparing with the convergent geometric series $2\sum_{n=0}^\infty
\left( \frac{2}{3}\right)^n$ yields that $\sum_{n=1}^\infty
(n+2^n) / (n+3^n)$ converges.

\item {\bf converges absolutely:} since $1/( n^2 \ln(n)) &lt; 1/n^2$ for
$n\ge 3$, since $\ln(n)\ge 1$ for $n\ge 3$, we have by the second
comparison test that $\sum_{n=2}^\infty 1/( n^2 \ln(n))$
converges.

\item {\bf diverges:} for large values of $n$, it seems that the
$n^{th}$ in the series is approximately $\frac{1}{n}$, and so we
might guess that the series diverges by the limit comparison test.
To check this guess, we need to evaluate
\[ \lim_{n\rightarrow\infty} \frac{ (n^3+1) / (n^4+2)}{1/n}
=\lim_{n\rightarrow\infty} \frac{ n^4+n}{n^4+2} = 1 = L. \] Since
the limit exists and since $0 &lt; L = 1&lt; \infty$, and since the
harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, we have
that $\sum_{n=1}^\infty (n^3+1) / (n^4+2)$ diverges by the limit
comparison test.

\item {\bf converges absolutely:} since $\frac{1}{n + n^{3/2}} &lt;
\frac{1}{n^{3/2}}$ for all $n\ge 1$ and since $\sum_{n=1}^\infty
\frac{1}{n^{3/2}}$ converges by note 1., we have that
$\sum_{n=1}^\infty 1 / (n + n^{3/2})$ converges by the second
comparison test.

\item {\bf converges absolutely:} for large values of $n$, it seems
that the $n^{th}$ term in this series is approximately equal to
$\frac{10}{n^2}$, and so we might guess that this series converges
by use of the limit comparison test.  To verify this guess, we
calculate
\[ \lim_{n\rightarrow\infty} \frac{10 n^2 / (n^4+1)}{10/n^2}
=\lim_{n\rightarrow\infty} \frac{n^4}{n^4+1} =1 =L. \] Since the
limit exists and since $0 &lt; L=1 &lt;\infty$, and since
$\sum_{n=1}^\infty \frac{10}{n^2}$ converges by note 1., we have
that $\sum_{n=1}^\infty 10 n^2 / (n^4+1)$ converges by the limit
comparison test.

\item {\bf converges absolutely:} for large values of $n$, it seems
again that the $n^{th}$ term in this series is approximately equal
to $\frac{1}{n^2}$, and so we might guess that this series
converges by use of the limit comparison test.  To verify this
guess, we calculate
\[ \lim_{n\rightarrow\infty} \frac{(n^2 -n) / (n^4 +2)}{1/n^2}
=\lim_{n\rightarrow\infty} \frac{n^4 -n^3}{n^4+2} =1 =L. \] Since
the limit exists and since $0 &lt; L=1 &lt;\infty$, and since
$\sum_{n=1}^\infty \frac{1}{n^2}$ converges by note 1., we have
that $\sum_{n=2}^\infty (n^2 -n) / (n^4 +2)$ converges by the
limit comparison test.

\item {\bf diverges:} for large values of $n$, it seems that the
$n^{th}$ term of this series is approximately equal to
$\frac{1}{n}$, and so we might guess that this series then
diverges by the limit comparison test.  To verify this guess, we
calculate
\[ \lim_{n\rightarrow\infty} \frac{1 / \sqrt{n^2+1}}{1/n}
=\lim_{n\rightarrow\infty} \frac{n}{\sqrt{n^2 +1}}  =
 \lim_{n\rightarrow\infty} \frac{n}{n\sqrt{ 1 + 1/n^2}} = 1 = L. \]
Since the limit exists and since $0 &lt; L=1 &lt;\infty$, and since
$\sum_{n=1}^\infty \frac{1}{n}$ diverges by note 1., we have that
$\sum_{n=2}^\infty 1/\sqrt{n^2 +1}$ diverges by the limit
comparison test.

\item {\bf converges absolutely:} since
\[ \frac{1}{3+5^n} &lt; \frac{1}{5^n} =\left( \frac{1}{5}\right)^n, \]
and since $\sum_{n=0}^\infty \left( \frac{1}{5}\right)^n$
converges, the second comparison test yields that
$\sum_{n=1}^\infty 1 / (3+5^n)$ converges.

\item {\bf diverges:} first note that since $\ln(n) &lt;n$ for all $n\ge
2$, this is a series of positive terms.  Also, $n -\ln(n) &lt;n$, and
so $1/(n -\ln(n)) &gt;1/n$.  Hence, since $\sum_{n=1}^\infty
\frac{1}{n}$ diverges, we have that $\sum_{n=2}^\infty 1 /
(n-\ln(n))$ diverges, by the first comparison test.

\item {\bf converges absolutely:} since $0 &lt;\cos^2(n) \le 1$ for all
$N\ge 1$, we have that $\cos^2(n) / 3^n &lt; 1/3^n$.  Since
$\sum_{n=0}^\infty \frac{1}{3^n}$ is a convergent geometric
series, we have by the second comparison test that
$\sum_{n=1}^\infty \cos^2(n) / 3^n$ converges.

\item {\bf converges absolutely:} since $1 / (2^n+3^n) &lt; 1/2^n$ and
since $\sum_{n=0}^\infty \frac{1}{2^n}$ converges, the second
comparison test yields that $\sum_{n=1}^\infty 1 / (2^n+3^n)$
converges.

\item {\bf converges absolutely:} since $1 + \sqrt{n}\ge 2$ for $n\ge
1$, we have that $n^{1+\sqrt{n}} \ge n^2$ for $n\ge 1$, and so $1
/ n^{(1+\sqrt{n})}\le 1/n^2$ for $n\ge 1$.  Hence, since
$\sum_{n=1}^\infty \frac{1}{n^2}$ converges by note 1., we have by
the second comparison test that $\sum_{n=1}^\infty 1 /
n^{(1+\sqrt{n})}$ converges.

\item {\bf converges absolutely:} since $2^n (n+1) &gt; 2^n$ for $n\ge
1$, we have that $1 / (2^n (n+1)) &lt;1/2^n$ for $n\ge 1$.  Since
$\sum_{n=1}^\infty \frac{1}{2^n}$ is a convergent geometric
series, we have by the second comparison test that
$\sum_{n=1}^\infty 1 / (2^n (n+1))$ converges.

\item {\bf diverges:} since factorials are involved, we first see
whether the ratio test gives us any information, and so we
evaluate
\[ \lim_{n\rightarrow\infty} \frac{(n+1)! / ((n+1)^2 e^{n+1})}{n!/(n^2
e^n)} =\lim_{n\rightarrow\infty} \frac{(n+1)!n^2 e^n}{n! (n+1)^2
e^{n+1}} = \lim_{n\rightarrow\infty} \frac{n^2}{(n+1)^2}
\frac{n+1}{e} =\infty, \] and since $\infty &gt;1$, the ratio test
implies that $\sum_{n=1}^\infty n! / (n^2 e^n)$ diverges.

\medskip
\noindent [Though it's not obvious how, we could also have applied
the $n^{th}$ term test for divergence, since for large values of
$n$ we have
\[ \frac{n!}{n^2 e^n} = \frac{(n-1)(n-2)!}{n e^n}
=\frac{n-1}{n}\frac{n-2}{e}\cdots \frac{2}{e} \frac{1}{e^2} &gt;
\frac{n-1}{n}\frac{2}{e} \frac{1}{e^2} &gt; \frac{1}{e^3}. \] We
simplified by noting that the middle terms $\frac{n-2}{e},\ldots,
\frac{3}{e}$ are all greater than $1$ and that $\frac{n-1}{n}
&gt;\frac{1}{2}$ for $n$ large.  Hence, $\lim_{n\rightarrow\infty}
\frac{n!}{n^2 e^n}\ne 0$.]

\item {\bf converges absolutely:} there is not an obvious comparison
to make, and so we try the ratio test:
\[ \lim_{n\rightarrow\infty} \frac{\sqrt{n+1}/ (3^{n+1}
\ln(n+1))}{\sqrt{n} / (3^n \ln(n))} =\lim_{n\rightarrow\infty}
\frac{1}{3} \frac{\ln(n)}{\ln(n+1)} \sqrt{\frac{n+1}{n}}
=\frac{1}{3}, \] since $\lim_{n\rightarrow\infty}
\frac{\ln(n)}{\ln(n+1)} =1$, for instance using l'Hopital's rule.
Since $\frac{1}{3} &lt;1$, the ratio test yields that
$\sum_{n=1}^\infty \sqrt{n} / (3^n \ln(n))$ converges.

\item {\bf converges absolutely:} since there are factorials involved,
we first try the ratio test:
\[ \lim_{n\rightarrow\infty} \frac{(2(n+1))!/((n+1)!)^3}{(2n)! /
(n!)^3} =\lim_{n\rightarrow\infty} \frac{(2n+2)(2n+1)}{(n+1)^3} =0
&lt; 1, \] and so the ratio test yields that $\sum_{n=2}^\infty (2n)!
/ (n!)^3$ converges.

\item {\bf converges absolutely:} note that the numerator of each term
is either $0$ or $2$, and so this is a series with non-negative
terms.  Also, $(1 - (-1)^n) / n^4 &lt; 2/n^4$ for all $n\ge 1$ and
$\sum_{n=1}^\infty \frac{1}{n^4}$ converges by note 1., and so by
the second comparison test $\sum_{n=1}^\infty (1 - (-1)^n) / n^4$
converges.

\item {\bf diverges:} we start with a bit of algebraic simplification:
\[ \frac{2+\cos(n)}{n + \ln(n)} \ge \frac{1}{n + \ln(n)} &gt;
\frac{1}{2n}. \] (The first inequality holds since $2+\cos(n) \ge
2 + (-1) = 1$ for all $n\ge 1$, and the second inequality holds
since $\ln(n) &lt; n$ for all $n\ge 1$, and so $n +\ln(n) &lt; n+n
=2n$.)  Since $\sum_{n=1}^\infty \frac{1}{2n}$ diverges (as it is
a constant multiple of the harmonic series), the first comparison
test yields that $\sum_{n=1}^\infty (2+\cos(n)) / (n + \ln(n))$
diverges.

\item {\bf diverges:} for this one, we use the integral test.  Set
\[ f(x) =\frac{1}{x \ln(x) \sqrt{\ln(\ln(x))}}, \]
so that $a_n =f(n)$ for all $n\ge 3$.  (The restriction that $n\ge
3$ is to ensure that $\sqrt{\ln(\ln(n))}$ is well defined.)  In
order to apply the integral test, we need to know that $f(x)$ is
decreasing, which involves calculating a derivative and checking
its sign:
\[ f'(x) =\frac{-\left( \ln(x)\sqrt{\ln(\ln(x))} + \sqrt{\ln(\ln(x))} +
x\ln(x) \frac{1}{2\sqrt{\ln(\ln(x))}}\frac{1}{x\ln(x)} \right)}{(x
\ln(x) \sqrt{\ln(\ln(x))})^2} &lt; 0. \] Hence, the integral test can
be applied, and says that $\sum_{n=3}^\infty 1 / (n \ln(n)
\sqrt{\ln(\ln(n))})$ converges if and only if $\int_3^\infty f(x)
{\rm d}x = \lim_{M\rightarrow\infty} \int_3^M f(x) {\rm d}x$
exists.   So, we calculate:
\[ \lim_{M\rightarrow\infty} \int_3^M f(x) {\rm d}x
=\lim_{M\rightarrow\infty} \int_3^M \frac{1}{x \ln(x)
\sqrt{\ln(\ln(x))}} {\rm d}x =\lim_{M\rightarrow\infty}
2\sqrt{\ln(\ln(x))}\left|_3^M \right., \] which diverges, and so
$\sum_{n=3}^\infty 1 / (n \ln(n) \sqrt{\ln(\ln(n))})$ diverges.

\item {\bf converges absolutely:} try the ratio test, since there are
factorials about:
\[ \lim_{n\rightarrow\infty} \frac{(n+1)^{(n+1)} / (\pi^{(n+1)}
(n+1)!)}{n^n / (\pi^n n!)} =\lim_{n\rightarrow\infty} \left(
\frac{n+1}{n}\right)^n \frac{1}{\pi} =\lim_{n\rightarrow\infty}
\left( 1 +\frac{1}{n}\right)^n \frac{1}{\pi} =\frac{e}{\pi} =L. \]
Since the limit exists and since $L &lt;1$, the ratio test yields
that $\sum_{n=1}^\infty n^n / (\pi^n n!)$ converges.

\item {\bf converges absolutely:} since both the numerator and the
denominator are raised to (essentially) the same power, we try the
root test, and so need to calculate:
\[ \lim_{n\rightarrow\infty} \left( \frac{2^{n+1}}{n^n}\right)^{1/n} =
\lim_{n\rightarrow\infty} 2^{1/n} \frac{2}{n} =L=0 \] (since
$\lim_{n\rightarrow\infty} 2^{1/n} =2^0 =1$).  Since the limit
exists and since $L &lt;1$, the root test yields that
$\sum_{n=1}^\infty 2^{n+1} / n^n$ converges.

\item {\bf converges conditionally:} we first test for absolute
convergence, by considering the related series $\sum_{n=1}^\infty
| (-1)^{n-1} / \sqrt{n}| =\sum_{n=1}^\infty 1 / \sqrt{n}$, which
diverges by note 1.

\medskip
\noindent We now test for convergence.  This is an alternating
series, and so we use the alternating series test: write
\[ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{\sqrt{n}} =(-1)
\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}} = (-1) \sum_{n=1}^\infty
(-1)^n a_n, \] where $a_n =\frac{1}{\sqrt{n}} &gt;0$ for all $n\ge
1$.  Since $\lim_{n\rightarrow\infty} a_n
=\lim_{n\rightarrow\infty} \frac{1}{\sqrt{n}} =0$ and since
$a_{n+1} =\frac{1}{\sqrt{n+1}} &lt;\frac{1}{\sqrt{n}} =a_n$ for all
$n\ge 1$, the alternating series test applies and yields that this
series converges.

\medskip
\noindent Hence, this series converges but does not converge
absolutely.  That is, the series converges conditionally.

\item {\bf converges conditionally:} we first check for absolute
convergence, that is, convergence of the associated series
$\sum_{n=1}^\infty |\cos(\pi n) / ( (n+1) \ln(n+1) ) | =
\sum_{n=1}^\infty 1 / ( (n+1) \ln(n+1) )$.  For this series, we
apply the integral test, with $f(x) = 1 / ( (x+1) \ln(x+1) )$.
Since
\[ f'(x) =\frac{-\left( \ln(x+1) + (x+1)\frac{1}{x+1}\right) }{(x+1)^2
(\ln(x+1))^2} =\frac{-( \ln(x+1) + 1) }{(x+1)^2(\ln(x+1))^2} &lt; 0
\] for $x\ge 1$, the integral test yields that the series
converges if and only if $\int_1^\infty f(x) {\rm d}x =
\lim_{M\rightarrow\infty} \int_1^M f(x) {\rm d}x$ exists, so we
calculate:
\[ \lim_{M\rightarrow\infty} \int_1^M \frac{1}{(x+1)\ln(x+1)} {\rm
d}x =\lim_{M\rightarrow\infty} \ln(\ln(x+1))\left|_1^M \right.
=\lim_{M\rightarrow\infty} (\ln(\ln(M+1)) -\ln(\ln(2))), \] which
diverges (very very slowly).  So, the series does not converge
absolutely.

\medskip
\noindent We now test for convergence.  Since $\cos(\pi n)
=(-1)^n$, this is an alternating series, and we start with the
alternating series test. Since $(n+1) \ln(n+1) &lt; (n+2) \ln(n+2)$
for all $n\ge 1$, we have that $1  / ( (n+1) \ln(n+1) ) &gt; 1/ (
(n+2) \ln(n+2) )$ for $n\ge 1$.  Since $\lim_{n\rightarrow\infty}
1/ ( (n+1) \ln(n+1) ) =0$ (and since  $1 / ( (n+1) \ln(n+1) ) &gt;0$
for $n\ge 1$), the alternating series test applies and yields that
the series converges.

\medskip
\noindent Hence, this series converges but does not converge
absolutely.  That is, the series converges conditionally.

\item {\bf diverges:} since $\lim_{n\rightarrow\infty} (n^2 -1) / (
n^2+1) =1$, we have that $\lim_{n\rightarrow\infty} (-1)^n (n^2
-1) / ( n^2+1)$ does not exist, and so $\sum_{n=1}^\infty (-1)^n
(n^2 -1) / ( n^2+1)$ diverges by the $n^{th}$ term test for
divergence.

\item {\bf converges absolutely:} we first test for absolute
convergence, by considering the associated series
$\sum_{n=1}^\infty |(-1)^n / (n \pi^n)| =\sum_{n=1}^\infty 1 / (n
\pi^n)$.  Since $1/(n\pi^n)\le 1/\pi^n$ for $n\ge 1$ and since
$\sum_{n=0}^\infty \frac{1}{\pi^n}$ converges, the second
comparison test yields that $\sum_{n=1}^\infty 1 / (n \pi^n)$
converges, and hence that $\sum_{n=1}^\infty (-1)^n / (n \pi^n)$
converges absolutely.

\item {\bf converges conditionally:} we first test for absolute
convergence, that is, convergence of the associated series
$\sum_{n=1}^\infty |(-1)^n (20n^2 -n -1) / (n^3+n^2+33 )|
=\sum_{n=1}^\infty (20n^2 -n -1) / (n^3+n^2+33 )$.  Since the
$n^{th}$ term looks like a constant multiple of $\frac{1}{n}$ for
large $n$, let's try the limit comparison test:
\[ \lim_{n\rightarrow\infty} \frac{(20n^2 -n -1) / (n^3+n^2+33)}{1/n}
=\lim_{n\rightarrow\infty} \frac{20n^3 -n^2 -n}{n^3+n^2+33} =20
=L. \] Since the limit exists and $0 &lt;L &lt;\infty$, the series being
considered here diverges, since the harmonic series converges. So,
the original series does not converge absolutely.

\medskip
\noindent We now test for convergence.  The series
$\sum_{n=1}^\infty (-1)^n (20n^2 -n -1) / (n^3+n^2+33 )
=\sum_{n=1}^\infty (-1)^n a_n$  is an alternating series, since
$\frac{20n^2 -n-1}{n^3 +n^2+33} &gt;0$ for $n\ge 1$, and so let's
check whether it satisfies the conditions of the alternating
series test.  Since $(20n^2 -n -1) / (n^3+n^2+33 )$ is a rational
function and the denominator has higher degree than the numerator,
we have that $\lim_{n\rightarrow\infty} (20n^2 -n -1) /
(n^3+n^2+33 ) =0$.  All that remains to check is whether the $a_n$
are monotonically decreasing.  For this, let $f(x) = (20x^2 -x -1)
/ (x^3+x^2+33 )$, so that $f(n) =a_n$, and check that it's
decreasing, which involves calculating $f'(x)$:
\[ f'(x) =\frac{-20x^4 +2x^3 +4x^2 +1322x -33}{(x^3 +x^2 +33)^2} &lt;0
\]
for all $x$ greater than any of the roots of the numerator.  So,
the alternating series test applies, and yields that this series
converges.

\medskip
\noindent Hence, this series converges but does not converge
absolutely.  That is, the series converges conditionally.

\item {\bf diverges:} note that, for $n\ge 101$, we have
\[ \frac{n!}{100^n} = \frac{ n(n-1)\cdots 1}{100^n} =
\frac{n}{100}\frac{n-1}{100}\cdots \frac{101}{100}\frac{100}{100}
\frac{99}{100}\cdots \frac{1}{100} &gt; \frac{99}{100}\cdots
\frac{1}{100}, \] and so $\lim_{n\rightarrow\infty}  n! /
(-100)^n$ does not exist. Hence, by the $n^{th}$ term test for
divergence, the series diverges.

\item {\bf converges absolutely:} we apply the integral test, with the
function $f(x) =\frac{1}{x \ln(x) (\ln(\ln(x)))^2}$.  First, we
check to see that $f(x)$ is decreasing, by calculating its
derivative:
\[ f'(x) =\frac{-(\ln(x) (\ln(\ln(x)))^2 + (\ln(\ln(x)))^2 + 1)}{(x
\ln(x) (\ln(\ln(x)))^4} &lt; 0 \] for $x\ge 2$ (and the denominator
is non-zero for $x\ge 3$). So, now we need to calculate
\begin{eqnarray*} \int_3^\infty f(x) {\rm d}x &amp; = &amp;
\lim_{M\rightarrow\infty} \int_3^M \frac{1}{x\ln(x)
(\ln(\ln(x))^2} {\rm d}x \\
 &amp; = &amp; \lim_{M\rightarrow\infty} \frac{1}{\ln(\ln(x)} \left|_3^M \right.
=\lim_{M\rightarrow\infty} \left( \frac{-1}{\ln(\ln(M))} +
\frac{1}{\ln(\ln(3))} \right) =\frac{1}{\ln(\ln(3))}.
\end{eqnarray*}
Since the limit converges, $\sum_{n=3}^\infty 1 / (n \ln(n)
(\ln(\ln(n)))^2)$ converges absolutely.

\item {\bf diverges:} we start with a bit of arithmetic, noting that
the numerator satisfies: $(1 + (-1)^n) =0$ for $n$ odd and $(1 +
(-1)^n) =2$ for $n$ even.  Hence, the terms of the series are
non-zero only for $n$ even, so let's make the substitution $n =2k$
for $k\ge 1$.  Then, for $n$ even, we have that
\[ \frac{1 + (-1)^n}{\sqrt{n}} =\frac{2}{\sqrt{2k}} =
\frac{\sqrt{2}}{\sqrt{k}} &gt; \frac{1}{\sqrt{k}}. \] Hence, by the
first comparison test and note 1., we have that $\sum_{n=1}^\infty
(1 + (-1)^n) / \sqrt{n}$ diverges.
\item {\bf converges absolutely:} again, we begin with a bit of
algebra, simplifying the $n^{th}$ in the series by noting that
\[ \frac{e^n \cos^2(n)}{1+\pi^n} \le \frac{e^n}{1+\pi^n} \le
\frac{e^n}{\pi^n} =\left( \frac{e}{\pi}\right)^n, \] where the
first inequality follows from $\cos^2(n)\le 1$ for all $n\ge 1$.
Since $\sum_{n=0}^\infty (e/\pi)^n$ converges, the second
comparison test yields that $\sum_{n=1}^\infty e^n \cos^2(n) /
(1+\pi^n)$ converges.

\item {\bf converges absolutely:} since there are factorials involved,
let's first try the ratio test:
\[ \lim_{n\rightarrow\infty} \frac{(n+1)^4 / (n+1)!}{n^4/n!}
=\lim_{n\rightarrow\infty} \left( \frac{n+1}{n}\right)^4
\frac{n!}{(n+1)!} ==\lim_{n\rightarrow\infty} \left(
\frac{n+1}{n}\right)^4 \frac{1}{n+1} =0 =L. \] Since the limit
exists and since $L &lt;1$, the ratio test yields that the series
converges.

\item {\bf converges absolutely:} again, since there are factorials
involved, we first try the ratio test:
\[ \lim_{n\rightarrow\infty} \frac{(2(n+1))! 6^{(n+1)} / (3(n+1))!}{
(2n)! 6^n / (3n)!} =\lim_{n\rightarrow\infty}
\frac{6(2n+2)(2n+1)}{(3n+3)(3n+2)(3n+1)} =0 =L. \] Since the limit
exists and since $L &lt;1$, the ratio test yields that the series
converges.

\item {\bf converges absolutely:} and yet again, since there are
factorials involved, our first attempt should be with the ratio
test:
\[ \lim_{n\rightarrow\infty} \frac{ (n+1)^{100} 2^{(n+1)} /
\sqrt{(n+1)!}}{n^{100} 2^n / \sqrt{n!}} =\lim_{n\rightarrow\infty}
\left( \frac{n+1}{n}\right)^{100} \frac{2}{\sqrt{n+1}} =0 =L. \]
Since the limit exists and since $L &lt;1$, the ratio test yields
that this series converges.

\item {\bf diverges:} since there are factorials involved, we first
try the ratio test:
\[ \lim_{n\rightarrow\infty} \frac{ (1+(n+1)!) / (1+(n+1))!}{ (1+n!) /
(1+n)!} =\lim_{n\rightarrow\infty} \frac{1 +(n+1)!}{(1+n!)(n+2)}
=\lim_{n\rightarrow\infty} \frac{1/n! + n+1}{(1/n! +1)(n+2)} =1,
\] and so the ratio test gives no information.  (This discussion
was put in to remind you that the ratio test doesn't always work
with factorials.)

\medskip
\noindent Hmm.  Notice that when $n$ is large, $1 +n!$ is very
nearly equal to $n!$, and so $(1+n!)/(n+1)!$ is very nearly equal
to $n!/(n+1)! =1/(n+1)$.  So, let's try the limit comparison test
with $1/(n+1)$:
\[ \lim_{n\rightarrow\infty} \frac{(1+n!) / (1+n)!}{1/(n+1)}
=\lim_{n\rightarrow\infty} \frac{(n+1)(1+n!)}{(n+1)!}
=\lim_{n\rightarrow\infty} \frac{1+n!}{n!} =1=L. \] Since the
limit exists and since $\sum_{n=0}^\infty 1/(n+1)$ diverges (as
it's the harmonic series less the leading term), the series
$\sum_{n=3}^\infty (1+n!) / (1+n)!$ diverges by the limit
comparison test.
\item {\bf diverges:} again, since there are factorials involved, we
first try the ratio test:
\[ \lim_{n\rightarrow\infty} \frac{2^{2(n+1)} ((n+1)!)^2}{(2(n+1))!}
\frac{(2n)!}{2^{2n} (n!)^2} =\lim_{n\rightarrow\infty}
\frac{4(n+1)^2}{(2n+1)(2n+2)} =1, \] and so the ratio test yields
no information.

\medskip
\noindent So, let's explicitly try the $n^{th}$ term test for
divergence.  We start with a bit of algebraic massage, namely:
\[ 2^{2n} (n!)^2 =(2^n\cdot n!)^2 = ( (2n)\cdot (2n-2)\cdot
(2n-4)\cdots 4\cdot 2)^2, \] and so
\[ \frac{2^{2n}(n!)^2}{(2n)!} =\frac{ (2n)\cdot (2n) \cdot (2n-2)\cdot
(2n-2) \cdots 2\cdot 2}{(2n)\cdot (2n-1)\cdot (2n-2)\cdot (2n-3)
\cdots 2\cdot 1} = \frac{ (2n)\cdot (2n-2)\cdots 2}{(2n-1)\cdot
(2n-3) \cdots 1} &gt; 1. \] In particular, the limit
$\lim_{n\rightarrow\infty} 2^{2n} (n!)^2 / (2n)!$ cannot be zero,
and so the $n^{th}$ term test yields that $\sum_{n=1}^\infty
2^{2n} (n!)^2 / (2n)!$ diverges.
\item {\bf converges absolutely:} we first check for absolute
convergence, namely the convergence of the series
$\sum_{n=1}^\infty |(-1)^n / ( n^2 + \ln(n) )| =\sum_{n=1}^\infty
1 / ( n^2 + \ln(n) )$.  Since $n^2 +\ln(n) &gt; n^2$, we have that
$1/(n^2 +\ln(n)) &lt; 1/n^2$, and so by the second comparison test,
the series $\sum_{n=1}^\infty 1 / ( n^2 + \ln(n) )$ converges.
That is, the original series $\sum_{n=1}^\infty (-1)^n / ( n^2 +
\ln(n) )$ converges absolutely.

\item {\bf converges absolutely:} we begin with a bit of algebraic
massage, noting that
\[ \sum_{n=1}^\infty \frac{(-1)^{2n}}{2^n}  =\sum_{n=1}^\infty
\frac{((-1)^2)^n }{2^n}  = \sum_{n=1}^\infty \frac{1}{2^n}
=\sum_{n=1}^\infty \left( \frac{1}{2}\right)^n. \] This is a
convergent geometric series, converging to
\[ \frac{1}{1-\frac{1}{2}} -1 = 1. \]
(The subtraction of $1$ arises from the fact that the starting
index in this series is not $0$, so that
\[ \sum_{n=1}^\infty \frac{1}{2^n} =\sum_{n=0}^\infty \frac{1}{2^n} -
\left( \frac{1}{2}\right)^0 =  \sum_{n=0}^\infty \frac{1}{2^n} - 1
= 2-1 = 1.) \]

\item {\bf converges absolutely:} we first check for absolute
convergence, namely the convergence of the series
$\sum_{n=1}^\infty |(-2)^n / n!| =\sum_{n=1}^\infty 2^n / n!$.
Since there are factorials involved, we make use of the ratio
test:
\[ \lim_{n\rightarrow\infty} \frac{2^{(n+1)} / (n+1)!}{2^n/n!}
=\lim_{n\rightarrow\infty} \frac{2}{n+1} =0 =L. \] Since this
limit exists and satisfies $L &lt;1$, the ratio test yields that
$\sum_{n=1}^\infty 2^n / n!$ converges, and hence that the
original series $\sum_{n=1}^\infty (-2)^n / n!$ converges
absolutely.

\item {\bf diverges:} first, note that this is not an alternating
series, but is a series with all non-positive terms.  Hence, for
this series, convergence and absolute convergence are equivalent,
as they are for series with non-negative terms.

\medskip
\noindent Now, for $n$ large, $n/(n^2 +1)$ is approximately equal
to $1/n$, and so let's try the limit comparison test with
$\frac{1}{n}$.  So, we calculate:
\[ \lim_{n\rightarrow\infty} \frac{ n / (n^2+1)}{1/n}
=\lim_{n\rightarrow\infty} \frac{ n^2}{n^2+1} =1 =L. \] Since the
limit exists and since $0 &lt; L =1 &lt;\infty$, and since
$\sum_{n=1}^\infty -1/n$ diverges (as it is a constant multiple of
the harmonic series), the limit comparison test yields that the
series $\sum_{n=1}^\infty -n / (n^2+1)$ diverges.

\item {\bf converges conditionally:} we start by noting that
$\cos(n\pi) =(-1)^n$, and so this is an alternating series.  So,
we first check for absolute convergence, namely the convergence of
the series $\sum_{n=1}^\infty |100\cos(n\pi) / (2n+3)| =
\sum_{n=1}^\infty 100 / (2n+3)$.  Here, there are many tests that
yield divergence, for instance we may use the limit comparison
test with the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$:
\[ \lim_{n\rightarrow\infty} \frac{100/(2n+3)}{1/n}
=\lim_{n\rightarrow\infty} \frac{100n}{2n+3} =50 =L; \] since this
limit exists and satisfies $0 &lt; L =50 &lt;\infty$, and since the
harmonic series diverges, the limit comparison test yields that
$\sum_{n=1}^\infty 100 / (2n+3)$ diverges.

\medskip
\noindent However, since $\frac{100}{2(n+1) +3} =\frac{100}{2n+5}
&lt; \frac{100}{2n+3}$ and since $\lim_{n\rightarrow\infty}
\frac{100}{2n+3} =0$, the alternating series test yields that
$\sum_{n=1}^\infty 100\cos(n\pi) / (2n+3)$ converges.

\medskip
\noindent Hence, this series converges but does not converge
absolutely.  That is, the series converges conditionally.

\item {\bf converges conditionally:} as before, we begin by
simplifying the expression of each term.  Here, note that
$\sin((n+1/2)\pi) =(-1)^n$, and so this is an alternating series.
As always, we first check for absolute convergence, namely the
convergence of the series $\sum_{n=10}^\infty |\sin((n+1/2)\pi) /
\ln(\ln(n))| =\sum_{n=10}^\infty 1 / \ln(\ln(n))$.  Since $n &gt;
\ln(\ln(n))$ for all $n\ge 10$, we have that $1/\ln(\ln(n)) &gt; 1/n$
for all $n\ge 10$, and so the series $\sum_{n=10}^\infty 1/
\ln(\ln(n))$ diverges by the first comparison test.  That is, the
original series does not converge absolutely.

\medskip
\noindent We are now ready to determine convergence of the
original series.  As this is an alternating series, let's check
whether the hypthoses of the alternating series test are
satisfied.  Since $1/\ln(\ln(n)) &gt; 1/\ln(\ln(n+1))$ and since
$\lim_{n\rightarrow\infty} 1/\ln(\ln(n)) =0$ (since
$\lim_{n\rightarrow\infty} \ln(\ln(n)) =\infty$), the alternating
series test applies to this series, and yields that the series
$\sum_{n=10}^\infty \sin((n+1/2)\pi) / \ln(\ln(n))$ converges.

\medskip
\noindent Hence, this series converges but does not converge
absolutely.  That is, the series converges conditionally.

\item {\bf diverges:} similar to the algebraic manipulation we
performed on the series whose terms were the reciprocals of the
terms in this series, we calculate:
\begin{eqnarray*}
\frac{(2n)!}{2^{2n} (n!)^2} &amp; = &amp; \frac{(2n)!}{(2^n n!)^2} \\
 &amp; = &amp; \frac{(2n)\cdot (2n-1)\cdot (2n-2)\cdot (2n-3) \cdots 2\cdot
 1}{(2n)\cdot  (2n) \cdot (2n-2) \cdot (2n-2)\cdots 2\cdot 2} \\
 &amp; = &amp; \frac{(2n-1)\cdot (2n-3)\cdots 3\cdot 1}{(2n)  \cdot
 (2n-2)\cdots 4\cdot 2 } \\
 &amp; = &amp; \frac{1}{2n} \frac{2n-1}{2n-2} \frac{2n-3}{2n-4} \cdots
 \frac{5}{4} \frac{3}{2} &gt; \frac{1}{2n}.
\end{eqnarray*}
Hence, since the series $\sum_{n=1}^\infty \frac{1}{2n}$ diverges
(as it is a constant multiple of the harmonic series), the first
comparison test yields that $\sum_{n=1}^\infty (2n)! / ( 2^{2n}
(n!)^2)$ diverges.

\item {\bf converges absolutely:} since each term is a power, we first
attempt to apply the root test, and so we calculate:
\[ \lim_{n\rightarrow\infty} \left[ \left( \frac{n}{n+1} \right)^{n^2}
\right]^{1/n} = \lim_{n\rightarrow\infty} \left( \frac{n}{n+1}
\right)^n =\lim_{n\rightarrow\infty} \left(
\frac{n+1}{n}\right)^{-n} = \frac{1}{\lim_{n\rightarrow\infty}
\left( 1 +\frac{1}{n}\right)^n} = \frac{1}{e} =L. \] Since the
limit exists and since $L &lt; 1$, the root test yields that
$\sum_{n=1}^\infty (n / (n+1) )^{n^2}$ converges.

\item {\bf converges absolutely:} we begin with a bit of algebraic
manipulation, namely noting that
\[ 1+2+\cdots+n =\frac{n(n+1)}{2} \]
for $n\ge 1$, and so
\[ \frac{1}{1+2+\cdots+n} =\frac{2}{n(n+1)} &lt; \frac{2}{n^2} \]
for $n\ge 1$.  Since $\sum_{n=1}^\infty 1/n^2$ converges, by note
1., the second comparison test yields that $\sum_{n=1}^\infty 1 /
(1+2+\cdots+n)$ converges.

\item {\bf converges absolutely:} we begin with a bit of
simplification, namely noting that
\[ 0\le \frac{\ln(n)}{2n^3 - 1} \le \frac{n}{2n^3 -1}\le
\frac{n}{n^3} = \frac{1}{n^2} \] for $n\ge 1$.  (The first
inequality follows since $\ln(n)\le n$ for $n\ge 1$, while the
second inequality follows since $2n^3 -1 \ge n^3$ for $n\ge 1$.)
Since $\sum_{n=1}^\infty 1/n^2$ converges by note 1., the second
comparison test yields that $\sum_{n=1}^\infty \ln(n) / (2n^3 -
1)$ converges.

\item {\bf converges absolutely:} note that this is not an alternating
series, even though the terms are not all of the same sign (since
$\sin(n)$ behaves a bit strangely).  However, we still begin
testing for convergence by testing for absolute convergence,
namely the convergence of the series $\sum_{n=1}^\infty |\sin(n) /
n^2|$.  Since $|\sin(n)|\le 1$ for all $n\ge 1$, and since
$\sum_{n=1}^\infty 1/n^2$ converges by note 1., the second
comparison test yields that $\sum_{n=1}^\infty \sin(n) / n^2$
converges absolutely.

\item {\bf diverges:} since $\lim_{n\rightarrow\infty} (n-1) / n =1$,
we have that $\lim_{n\rightarrow\infty} (-1)^n (n-1) / n$ does not
exist (since for large $n$, it is oscillating between numbers near
$1$ and numbers near $-1$).  Since this limit does not exist, the
$n^{th}$ term test for divergence yields that $\sum_{n=1}^\infty
(-1)^n (n-1) / n$ diverges.

\item {\bf diverges:} we can rewrite this series as a geometric
series, to whit:
\[ \sum_{n=1}^\infty \frac{(-1)^n 2^{3n}}{7^n} =\sum_{n=1}^\infty
\frac{(-8)^n}{7^n} = \sum_{n=1}^\infty \left(
\frac{-8}{7}\right)^n. \] Since $|-\frac{8}{7}| \ge 1$, this is a
divergent geometric series.

\item {\bf converges absolutely:} this is similar to a series we
handled a few problems ago.  Even though the terms are not of the
same sign and are not of alternating signs, we still begin our
check for convergence by checking for absolute convergence.  Since
$|\cos(n) / n^4| \le 1/n^4$ (since $|\cos(n)| \le 1$ for all $n\ge
1$) and since $\sum_{n=1}^\infty 1/n^4$ converges, the second
comparison test yields that $\sum_{n=1}^\infty \cos(n) / n^4$
converges absolutely.

\item {\bf diverges:} even though this is an alternating series, I
personally feel the need to try the $n^{th}$ term test first,
since for $n$ large, the dominant terms are the $3^n$ in the
numerator and the $2^n$ in the demoninator, and so I expect that
the value of $3^n / (n(2^n + 1))$ to be large for large values of
$n$.  Let's check this:
\[ \frac{3^n}{n(2^n + 1)} =\frac{3^n}{n\: 2^n + n} &gt; \frac{3^n}{n\:
2^n + n\: 2^n} =\frac{3^n}{2n\: 2^n} =\left(\frac{3}{2}\right)^n
\frac{1}{2n}. \] Now, notice that $(3/2)^n &gt; n$ for $n\ge 3$
(since $(3/2)^3 &gt;3$ and the derivative of $(3/2)^n -n$ is positive
for $n\ge 3$), and so
\[ \frac{3^n}{n(2^n + 1)} &gt; \left(\frac{3}{2}\right)^n \frac{1}{2n} &gt;
\frac{1}{2} \] for $n\ge 3$.  (So, not exactly large for large
values of $n$, but big enough to do the trick.)  Hence, the limit
$\lim_{n\rightarrow\infty} (-1)^n 3^n / (n(2^n + 1))$ does not
exist (as it oscillates positive and negative and never settles
down to $0$), and so by the $n^{th}$ term test for divergence,
$\sum_{n=1}^\infty (-1)^n 3^n / (n(2^n + 1))$ diverges.

\item {\bf converges conditionally:} we first check for absolute
convergence, namely the convergence of the series
$\sum_{n=1}^\infty |(-1)^{n-1} n / (n^2+1)| =\sum_{n=1}^\infty n /
(n^2+1)$.  Since $n/(n^2 +1) &gt; n/(n^2 + n^2) = 1/(2n)$ for all
$n\ge 1$ and since $\sum_{n=1}^\infty 1/(2n)$ diverges (as it is a
constant multiple of the harmonic series), the first comparison
test yields that $\sum_{n=1}^\infty  n / (n^2+1)$ diverges, and so
the original series does not converge absolutely.

\medskip
\noindent As it is an alternating series, we can attempt to check
convergence by seeing if we can apply the alternating series test.
Since $\lim_{n\rightarrow\infty} n/(n^2 +1) =0$ and since $n/(n^2
+1) &gt; (n+1)/((n+1)^2 +1)$ for all $n\ge 1$, the hypotheses of the
alternating series test are met, and so $\sum_{n=1}^\infty
(-1)^{n-1} n / (n^2+1)$ converges.

\medskip
\noindent Hence, this series converges but does not converge
absolutely.  That is, the series converges conditionally.
\item {\bf converges absolutely:} we first check absolute convergence,
namely the convergence of the series $\sum_{n=2}^\infty
|(-1)^{n-1} / (n\ln^2(n))| =\sum_{n=2}^\infty 1 / (n\ln^2(n))$.
For this series, we use the integral test: set $f(x) =1 /
(x\ln^2(x))$. We need to check that $f(x)$ is decreasing, which we
do by calculating its derivative:
\[ f'(x) =\frac{-(\ln^2(x) + 2\ln(x))}{x^2 \ln^4(x)} &lt;0 \]
for $x\ge 2$ (since $\ln(x) &gt; 0$ for $x\ge 2$).  We now calculate:
\begin{eqnarray*} \int_2^\infty f(x) {\rm d}x &amp; = &amp;
\lim_{M\rightarrow\infty} \int_2^M \frac{1}{x\ln^2(x)} {\rm d}x \\
 &amp; = &amp; \lim_{M\rightarrow\infty} \frac{-1}{\ln(x)} \left|_2^M \right.\\
 &amp; = &amp; \lim_{M\rightarrow\infty} \left( \frac{-1}{\ln(M)} +
 \frac{1}{\ln(2)}\right) =\frac{1}{\ln(2)}.
\end{eqnarray*}
Since this limit exists, the integral test yields that the series
$\sum_{n=2}^\infty 1 / (n\ln^2(n))$ converges, and hence that the
original series $\sum_{n=2}^\infty (-1)^{n-1} / (n\ln^2(n))$
converges absolutely.

\item {\bf diverges:} we apply the ratio test (note 2.), as this is a series with non-zero
terms:
\[ \lim_{n\rightarrow\infty} \left| \frac{(-1)^n 2^{(n+1)} /
(n+1)^2}{(-1)^{n-1} 2^n / n^2}\right| =\lim_{n\rightarrow\infty}
 \frac{2n^2}{(n+1)^2} =2 =L. \]
Since this limit exists and satisfies $L &gt;1$, the series
$\sum_{n=1}^\infty (-1)^{n-1} 2^n / n^2$ diverges.

\item {\bf converges absolutely:} we first check for absolute
convergence, namely the convergence of the series
$\sum_{n=1}^\infty |(-1)^n \sin(\sqrt{n}) / n^{3/2}|
=\sum_{n=1}^\infty |\sin(\sqrt{n})| / n^{3/2}$.  Since $
|\sin(\sqrt{n})| / n^{3/2} \le 1/ n^{3/2}$ for $n\ge 1$ (since
$|\sin(\sqrt{n})| \le 1$ for $n\ge 1$), and since
$\sum_{n=1}^\infty 1 / n^{3/2}$ converges by note 1., the second
comparison test yields that $\sum_{n=1}^\infty | \sin(\sqrt{n})| /
n^{3/2}$ converges, and hence that the original series
$\sum_{n=1}^\infty (-1)^n \sin(\sqrt{n}) / n^{3/2}$ converges
absolutely.

\item {\bf converges absolutely:} even though there are no factorials,
let us apply the ratio test.  So, we calculate:
\[ \lim_{n\rightarrow\infty} \frac{ (n+1)^4 e^{-(n+1)^2}}{n^4
e^{-n^2}} =\lim_{n\rightarrow\infty} \left( \frac{n+1}{n}\right)^4
e^{-2n-1} =0 =L. \] Since this limit exists and since $L &lt;1$, the
ratio test yields that the series $\sum_{n=1}^\infty n^4 e^{-n^2}$
converges.

\item {\bf converges conditionally:} before testing for absolute
convergence, we perform a bit of algebraic simplification, by
noting that
\[ \sin\left( \frac{n \pi}{2}\right) = \sin\left( \frac{2k\pi}{2}
\right)  = \sin(k\pi) =0\] for $n$ even and
\[ \sin\left( \frac{\pi n}{2} \right) =\sin\left( \frac{\pi (2k+1)}{2}
\right) = \sin\left( k\pi + \frac{\pi}{2}\right) =(-1)^k\] for $n
=2k+1$ odd.  Hence, setting $n =2k+1$ for $k\ge 0$, we may rewrite
the series as
\[ \sum_{n=1}^\infty \frac{\sin(n\pi /2)}{n} =\sum_{k=0}^\infty
\frac{\sin(\pi (2k+1)/2)}{2k+1} =\sum_{k=0}^\infty
\frac{(-1)^k}{2k+1}. \]

\medskip
\noindent We first test for absolute convergence, namely the
convergence of the series $\sum_{k=0}^\infty |(-1)^k/(2k+1)| =
\sum_{k=0}^\infty 1/(2k+1)$.  However, since $1/(2k+1) &gt; 1/(2k+2)
=1/2(k+1)$ and since $\sum_{k=0}^\infty 1/(k+1)$ is the harmonic
series, the series $\sum_{k=0}^\infty 1/(2k+1)$ diverges by the
first comparison test, and hence the original series does not
converge absolutely.

\medskip
\noindent To test convergence, we use the alternating series test.
Since $1/(2k+1) &gt; 1/(2(k+1) +1)$ for all $k\ge 0$ and since
$\lim_{k\rightarrow\infty} 1/(2k+1) =0$, the alternating series
test yields that $\sum_{k=0}^\infty (-1)^k/(2k+1)$ converges.

\medskip
\noindent Hence, this series converges but does not converge
absolutely.  That is, the series converges conditionally.
\item {\bf diverges:} for this series, we first note that $\ln(x) &lt;
x^{1/8}$ for $x$ large ($x &gt; e^{32}$ works), as follows: consider
the function $f(x) =x^{1/8} -\ln(x)$, and note that
\[ f(e^{8k}) = (e^{8k})^{1/8} -\ln(e^{8k}) = e^k -8k, \]
and so $f(e^{32}) =e^4 - 32 = 22.5982 ... &gt; 0$.

\medskip
\noindent Moreover, for $x\ge e^{32}$, we have that $f(x)$ is
increasing: differentiating, we see that
\[ f'(x) = \frac{1}{8} x^{-7/8} -\frac{1}{x} =\frac{1}{x} \left(
\frac{1}{8} x -1 \right), \] and so $f'(x) &gt;0$ for $x &gt; 8$.

\medskip
\noindent So, for $n &gt; e^{32}$, we have that
\[ \frac{1}{\ln(n)^8} &gt; \frac{1}{(n^{1/8})^8} =\frac{1}{n}, \]
and hence by the first comparison test, $\sum_{n=2}^\infty 1 /
(\ln(n))^8$ diverges.  (Note that we are making heavy use of Note
3., that ignoring finitely many terms of a series does not affect
its convergence or divergence.)

\item {\bf converges absolutely:} (note that the lower limit $13$ for
the series yields that $\ln(n)$ and $\ln(\ln(n))$ are positive for
all terms in the series.)  We apply the integral test, using the
function
\[ f(x) = \frac{ 1}{x\ln(x) (\ln(\ln(x)))^p}. \]
We first check that $f(x)$ is decreasing:
\[ f'(x) = \frac{-(\ln(x)\ln(\ln(x))^p + \ln(\ln(x))^p +
p)}{(x\ln(x)\ln(\ln(x))^p)^2} &lt; 0 \] for $x &gt;13$, since both
$\ln(x) &gt;0$ and $\ln(\ln(x))&gt;0$ for $x &gt;13$ and since $p &gt;0$ by
assumption.

\medskip
\noindent In order to apply the integral test, we now need to
calculate:
\[ \int_{13}^\infty f(x) {\rm d}x = \lim_{M\rightarrow\infty}
\int_{13}^M \frac{ 1}{x\ln(x) (\ln(\ln(x)))^p}. \] There are two
cases: if $p = 1$, we get
\begin{eqnarray*} \lim_{M\rightarrow\infty} \int_{13}^M \frac{
1}{x\ln(x) (\ln(\ln(x))} &amp; = &amp; \lim_{M\rightarrow\infty}
\ln(\ln(\ln(x))) \left|_{13}^M \right. \\
 &amp; = &amp; \lim_{M\rightarrow\infty} \left( \ln(\ln(\ln(M)))
-\ln(\ln(\ln(13))) \right) =\infty,
\end{eqnarray*}
and so for $p =1$ the series diverges.

\medskip
\noindent For $p\ne 1$, we get:
\begin{eqnarray*} \lim_{M\rightarrow\infty} \int_{13}^M \frac{
1}{x\ln(x) (\ln(\ln(x))^p } &amp; = &amp; \lim_{M\rightarrow\infty}
\frac{1}{-p+1} \frac{1}{\ln(\ln(x))^{p-1}} \left|_{13}^M \right.
\\
 &amp; = &amp; \frac{1}{-p+1}  \lim_{M\rightarrow\infty} \left(
\ln(\ln(M))^{-p+1} -\ln(\ln(13))^{-p+1} \right),
\end{eqnarray*}
which converges for $p &gt;1$ (since $-p+1 &lt;0$)and diverges for $p
&lt;1$ (since $-p +1 &gt;0$).  Hence, the series $\sum_{n=13}^\infty 1
/( n\ln(n) (\ln(\ln(n)))^p )$ converges if and only if $p &gt;1$.
(Note that this is really just Note 1. in a bit of disguise.)
\end{enumerate}

\noindent Note 1.

\noindent The series $\sum_{n=1}^\infty \frac{1}{n^s}$ converges
if and only if $s &gt;1$.

\medskip
\noindent For $s =1$, this series is called the {\bf harmonic
series}, and we can prove directly that it diverges.  Note that
$\frac{1}{3} + \frac{1}{4} &gt; \frac{1}{2}$, that $\frac{1}{5} +
\cdots + \frac{1}{8} &gt; 4\frac{1}{8} = \frac{1}{2}$, and in general
that
\[ \frac{1}{2^{k-1} +1} + \frac{1}{2^{k-1} +2} +\cdots +\frac{1}{2^k}
&gt; 2^{k-1}\frac{1}{2^k} =\frac{1}{2}. \]
Hence, the $(2^k)^{th}$ partial sum $S_{2^k}$ satisfies $S_{2^k}
&gt;1 +k\frac{1}{2}$.  Since the terms in the harmonic series are all
positive, the sequence of partial sums is monotonically
increasing, and by the calculation done the sequence of partial
sums is unbounded, and so the sequence of partial sums diverges.
Hence, the harmonic series diverges.

\bigskip

\noindent Note 2.

\noindent {\bf Ratio and root tests for general series:} Let
$\sum_{n=0}^\infty a_n$ be a series with non-zero terms, so that
$a_n\ne 0$ for all $n$.
\begin{itemize}
\item {\bf Ratio test:} Suppose that $\lim_{n\rightarrow\infty} \left|
\frac{a_{n+1}}{a_n}\right| = L$ exists.  If $L &lt;1$, then
$\sum_{n=0}^\infty a_n$ converges absolutely.  If $L &gt; 1$, then
$\sum_{n=0}^\infty a_n$ diverges.  If $L =1$, this test gives no
information.
\item {\bf Root test:} Suppose that $\lim_{n\rightarrow\infty} (| a_n|
)^{1/n} = L$ exists.  If $L &lt;1$, then $\sum_{n=0}^\infty a_n$
converges absolutely.  If $L &gt; 1$, then $\sum_{n=0}^\infty a_n$
diverges.  If $L =1$, this test gives no information.
\end{itemize}

\bigskip

\noindent Note 3.

\noindent Let $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$
be two infinite series, and suppose there exists $P$ so that $a_n
=b_n$ for all $n &gt;P$.  (That is, assume the terms of the two
series are equal after some point.)  Then, $\sum_{n=0}^\infty a_n$
converges if and only if $\sum_{n=0}^\infty b_n$ converges.  That
is, the convergence or divergence of a series is not affected by
mucking about with finitely many terms of the series.
\end{document}
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