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\noindent {\bf Question} {\bf The sequence scavenger hunt}: for
each of the following sequences $\{ a_n\}$, do the following:
\begin{itemize}
\item Determine whether the sequence converges or diverges;
\item if the sequence converges, determine its limit;
\item if the sequence diverges, determine whether the sequence
converges to $\infty$ or if the sequence converges to $-\infty$ or
neither;
\end{itemize}

\begin{enumerate}
\item $a_n= (n+2)^{1/n}$;
\item $a_n =\frac{n^2 + 3n + 2}{6n^3 + 5}$;
\item $a_n = (1+\frac{1}{n})^n$;
\item $a_n = \frac{\sin(n)}{3^n}$;
\item $a_n=\sqrt{2n+3} -\sqrt{n+1}$;
\item $a_n=\cos \left( \frac{n\pi}{4} \right)$;
\item $a_n=(1+\frac{1}{n})^{1/n}$;
\item $a_n =\ln(n)$;
\item $a_n =e^n$;
\item $a_n = \frac{\ln(n)}{\sqrt{n}}$;
\item $a_n = \left( 1-\frac{2}{n^2} \right)^n$;
\item $a_n = \frac{n^3}{10n^2+1}$;
\item $a_n =x^n$, where $x$ is a constant with $|x|<1$;
\item $a_n =\frac{c}{n^p}$, where $c\neq 0$ and $p>0$ are constants;
\item $a_n = \frac{2n}{5n-3}$;
\item $a_n =\frac{1-n^2}{2+3n^2}$;
\item $a_n = \frac{n^3-n+7}{2n^3+n^2}$;
\item $a_n =1 + \left( \frac{9}{10}\right)^n$;
\item $a_n =2- \left( -\frac{1}{2} \right)^n$;
\item $a_n =1+(-1)^n$;
\item $a_n = \frac{1+(-1)^n}{n}$;
\item $a_n = \frac{1+(-1)^n\sqrt{n}}{ \left( \frac{3}{2} \right)^n}$;
\item $a_n = \frac{\sin^2(n)}{\sqrt{n}}$;
\item $a_n =\sqrt{\frac{2+\cos(n)}{n}}$;
\item $a_n =n\sin(\pi n)$;
\item $a_n =n\cos(\pi n)$;
\item $a_n =\pi^{-\sin(n)/n}$;
\item $a_n =2^{\cos(\pi n)}$;
\item $a_n = \frac{\ln(2n)}{\ln(3n)}$;
\item $a_n = \frac{\ln^2(n)}{n}$;
\item $a_n =n\sin \left( \frac{1}{n} \right)$;
\item $a_n = \frac{\arctan(n)}{n}$;
\item $a_n = \frac{n^3}{e^{n/10}}$;
\item $a_n = \frac{2^n+1}{e^n}$;
\item $a_n = \frac{\sinh(n)}{\cosh(n)}$;
\item $a_n =(2n+5)^{1/n}$;
\item $a_n = \left( \frac{n-1}{n+1} \right)^n$;
\item $a_n =(0.001)^{-1/n}$;
\item $a_n =2^{(n+1)/n}$;
\item $a_n = \left( \frac{2}{n} \right)^{3/n}$;
\item $a_n =(-1)^n (n^2+1)^{1/n}$;
\item $a_n = \frac{ \left( \frac{2}{3} \right)^n}{ \left( \frac{1}{2}
\right)^n+\left( \frac{9}{10} \right)^n}$;
\end{enumerate}
\medskip

\noindent {\bf Answer}
\begin{enumerate}
\item {\bf converges:} whenever we are evaluating a limit in which the
variable (in this case $n$) appears in both the base and the
exponent, we follow the same basic procedure.  First use the
identity $x =\exp(\ln(x))$ to rewrite the term.  Here,
\[ a_n= (n+2)^{1/n} =\exp\left( \frac{\ln(n+2)}{n}\right). \]
Next, we check to see whether we are dealing with an indeterminate
form.  Since the limit $\lim_{n\rightarrow\infty}
\frac{\ln(n+2)}{n}$ has the indeterminate form
$\frac{\infty}{\infty}$, we may use l'Hopital's rule to evaluate
\[ \lim_{n\rightarrow\infty} \frac{\ln(n+2)}{n}
=\lim_{n\rightarrow\infty} \frac{1}{n+2} =0. \] Hence, $\{ a_n\}$
converges to $e^0 =1$.

\item {\bf converges:} there is a standard way of evaluating the limit
as $n\rightarrow\infty$ of a rational function in $n$ (where a
rational function is the quotient of two polynomials).  First,
locate the highest power of $n$ that appears in either the
numerator or the denominator, and then multiply both numerator and
denominator by its reciprocal.  Here, the highest power of $n$
that appears is $n^3$, and so we calculate
\[ a_n =\frac{n^2 + 3n + 2}{6n^3 + 5} =\frac{n^2 + 3n + 2}{6n^3 + 5}
\cdot \frac{\frac{1}{n^3}}{\frac{1}{n^3}} =\frac{\frac{1}{n} +
\frac{3}{n^2} + \frac{2}{n^3}}{6+ \frac{5}{n^3}}. \] We then use
several properties of limits: that the limit of a quotient is the
quotient of the limits, that the limit of a sum is the sum of the
limits, and that $\lim_{n\rightarrow\infty} \frac{1}{n} =0$. Here,
\[ \lim_{n\rightarrow\infty} a_n =\lim_{n\rightarrow\infty}
\frac{\frac{1}{n} + \frac{3}{n^2} + \frac{2}{n^3}}{6+
\frac{5}{n^3}} =\frac{0}{6} =0. \] Hence, $\{ a_n\}$ converges to
$0$.

\item {\bf converges:} as above, we first rewrite the term using $x
=\exp(\ln(x))$.  Here,
\[ a_n = \left( 1+\frac{1}{n} \right)^n =\exp \left( n\ln\left(
1+\frac{1}{n} \right) \right) =\exp\left( \frac{\ln\left( 1 +
\frac{1}{n}\right) }{\frac{1}{n}} \right). \] We then concentrate
on the exponent and check to see whether we are dealing with an
indeterminate form, which in this case we are, since both
$\lim_{n\rightarrow\infty} \ln(1+\frac{1}{n})$ and
$\lim_{n\rightarrow\infty} \frac{1}{n}$ are equal to $0$. Hence,
we may apply l'Hopital's rule to evaluate
\[ \lim_{n\rightarrow\infty} \frac{\ln \left( 1 + \frac{1}{n} \right)
}{\frac{1}{n}} =\lim_{n\rightarrow\infty} \frac{1}{1+ \frac{1}{n}}
=1. \] Hence, $\{ a_n\}$ converges to $e^1 =e$.

\item {\bf converges:} here we use the squeeze law.  Since $-1\le
\sin(n)\le 1$ for all $n$, we have that $-\frac{1}{3^n} \le
\frac{\sin(n)}{3^n} \le \frac{1}{3^n}$.  Since
$\lim_{n\rightarrow\infty} \frac{1}{3^n} =0$, we have that
$\lim_{n\rightarrow\infty} -\frac{1}{3^n} = 0$ as well, and so $\{
a_n\}$ converges to $0$.

\item {\bf diverges:} write
\[ a_n=(\sqrt{2n+3} -\sqrt{n+1})\cdot\frac{\sqrt{2n+3}
+\sqrt{n+1}}{\sqrt{2n+3} +\sqrt{n+1}} = \frac{n+2}{\sqrt{2n+3}
+\sqrt{n+1}}. \] We now massage algebraically, in order to
simplify:
\[ \frac{n+2}{\sqrt{2n+3} +\sqrt{n+1}}\ge \frac{n+2}{2 \sqrt{2n+3}} =
\frac{n+ \frac{3}{2} + \frac{1}{2}}{2 \sqrt{2(n+\frac{3}{2} )}} >
\frac{n+\frac{3}{2}}{2 \sqrt{2(n+\frac{3}{2} )}}
=\frac{1}{2\sqrt{2}}\sqrt{n+\frac{3}{2} }. \] Since
$\lim_{n\rightarrow\infty} \sqrt{n+\frac{3}{2} } =\infty$, we see
by the comparison test that $\lim_{n\rightarrow\infty} a_n
=\infty$, and so $\{ a_n\}$ diverges.

\item {\bf diverges:} for $n = 8k$, $a_{8k}=\cos \left(
\frac{8k\pi}{4} \right) = 1$, while for $n=8k+1$, $a_{8k+1}=\cos
\left( \frac{(8k+1)\pi}{4} \right) =\frac{1}{\sqrt{2}}$.  In
particular, $|a_{8k} -a_{8k+1}| = \frac{1}{\sqrt{2}}$, and so the
sequence fails the Cauchy criterion, and so diverges.

\item {\bf converges:} write $a_n= \left( 1+\frac{1}{n} \right)^{1/n}
=\exp\left( \frac{ \ln \left( 1+\frac{1}{n} \right)}{n} \right)$.
Since $\lim_{n\rightarrow\infty} \ln(1+\frac{1}{n} ) = 0$, we have
that $\lim_{n\rightarrow\infty} \frac{\ln(1+\frac{1}{n})}{n} = 0$
(by the squeeze law for instance, since $0\le \frac{
\ln(1+\frac{1}{n})}{n} \le \ln(1+ \frac{1}{n})$ for $n\ge 1$).
Hence, $\lim_{n\rightarrow\infty} \exp \left( \frac{
\ln(1+\frac{1}{n})}{n} \right) =e^0 =1$, and so $\{ a_n\}$
converges to $1$.

\item {\bf diverges:} given $\varepsilon >0$, we show that there
exists $M$ so that $a_n >\varepsilon$ for $n >M$.  Since $a_n
=\ln(n)$, this becomes $\ln(n) >\varepsilon$ for $n >M$.
Exponentiating both sides of $\ln(n) >\varepsilon$, we get that $n
>e^\varepsilon$ (and vice versa, that if $n >e^\varepsilon$, then
$\ln(n) >\varepsilon$, since $e^x$ is an increasing function), and
so we can take $M =e^\varepsilon$.

\item {\bf diverges:} very similar to the question just done.  Given
$\varepsilon >0$, we show that there exists $M$ so that $a_n
>\varepsilon$ for $n>M$.  Taking logs of both sides of $a_n =e^n
>\varepsilon$, we get that $n >\ln(\varepsilon)$.  So, we make take $M
=\ln(\varepsilon)$.

\item {\bf converges:} since $\lim_{n\rightarrow\infty} a_n$ has the
indeterminate form $\frac{\infty}{\infty}$ (as both
$\ln(n)\rightarrow\infty$ and $\sqrt{n}\rightarrow\infty$ as
$n\rightarrow\infty$), we may apply l'Hopital's rule to see that
\[ \lim_{n\rightarrow\infty} \frac{\ln(n)}{\sqrt{n}} =
=\lim_{n\rightarrow\infty} \frac{ \frac{1}{n} }{
\frac{1}{2\sqrt{n}} } =\lim_{n\rightarrow\infty}
\frac{2}{\sqrt{n}} =0. \] Hence, $\{ a_n\}$ converges to $0$.

\item {\bf converges:} as always, we first rewrite each term as
\[ a_n = \left( 1- \frac{2}{n^2} \right)^n =\exp \left (n\ln \left( 1-
\frac{2}{n^2} \right) \right) =\exp \left( \frac{ \ln \left( 1-
\frac{2}{n^2} \right)}{\frac{1}{n}} \right). \] As
$n\rightarrow\infty$, the exponent reveals itself to have the
indeterminate form $\frac{0}{0}$, and so we may evaluate using
l'Hopital's rule:
\[ \lim_{n\rightarrow\infty} \frac{ \ln \left( 1-\frac{2}{n^2} \right)
}{ \frac{1}{n} } = \lim_{n\rightarrow\infty} \frac{
\frac{1}{1-\frac{2}{n^2} }\cdot \frac{4}{n^3}}{\frac{-1}{n^2}}
=\lim_{n\rightarrow\infty} \frac{- \frac{4}{1 -\frac{2}{n^2} }}{n}
= 0. \] Hence, $\{ a_n\}$ converges to $e^0 =1$.

\item {\bf diverges:} we could use either l'Hopital's rule (since the
limit has the indeterminate form $\frac{\infty}{\infty}$) or the
standard trick for dealing with limits of rational functions
(multiply numberator and denominator by the reciprocal of the
highest power of $n$ appearing anywhere in the term), but instead
we massage algebraically:
\[ a_n = \frac{n^3}{10n^2+1} > \frac{n^3}{10n^2+ 10n^2} =
\frac{n}{20}. \] Since $\{ \frac{n}{20} \}$ diverges, the
comparison test gives that $\{ a_n\}$ diverges as well.

\item {\bf converges:} it is a reasonable guess that $\{ a_n =x^n\}$
converges to $0$, which by definition means that given
$\varepsilon
>0$, there exists $M$ so that $| x^n -0| =| x^n| <\varepsilon$ for $n
>M$.  For $x =0$, this is true, since $\{ x^n\}$ becomes the constant
sequence $\{ a_n =0\}$.  So, we can assume that $x\ne 0$.  Taking
$\ln$ of both sides of $|x^n| <\varepsilon$ and using that $|x^n|
=|x|^n$, we get that $n\ln(|x|) <\ln(\varepsilon)$, and so $n >
\frac{\ln(\varepsilon)}{\ln(|x|)}$.  (The direction of the
inequality changes since $|x| <1$ and so $\ln(|x|) <0$.)  Hence,
we may take $M =\frac{\ln(\varepsilon)}{\ln(|x|)}$.  [Then, if $n
>M =\frac{\ln(\varepsilon)}{\ln(|x|)}$, then $n\ln(|x|)
<\ln(\varepsilon)$, and exponentiating we get that $|x|^n
<\varepsilon$, as desired.)

\item {\bf converges:} recall that $n^p \ge n$ and that
$n\rightarrow\infty$ as $n\rightarrow\infty$, and so
$n^p\rightarrow\infty$ as $n\rightarrow\infty$.  Hence, $\{
\frac{1}{n^p} \}$ converges to $0$, and therefore $\{ a_n
=\frac{c}{n^p} \}$ converges to $c\cdot 0 =0$.

\item {\bf converges:} using the standard trick for rational
functions, write
\[ a_n =\frac{2n}{5n-3} =\frac{2n}{5n-3}\cdot\frac{\frac{1}{n} }{
\frac{1}{n} } =\frac{2}{5-\frac{3}{n} }. \] As
$n\rightarrow\infty$, $\frac{1}{n} \rightarrow 0$ and so $\{
a_n\}$ converges to $\frac{2}{5}$.

\item {\bf converges:} using the standard trick for rational
functions, write
\[ a_n =\frac{1-n^2}{2+3n^2}
=\frac{1-n^2}{2+3n^2}\cdot\frac{ \frac{1}{n^2} }{ \frac{1}{n^2} }
=\frac{ \frac{1}{n^2} -1}{ \frac{2}{n^2} +3}. \] As
$n\rightarrow\infty$, $\frac{1}{n^2} \rightarrow 0$ and so $\{
a_n\}$ converges to $-\frac{1}{3}$.

\item {\bf converges:} using the standard trick for rational
functions, write
\[ a_n =\frac{n^3-n+7}{2n^3+n^2} =\frac{n^3-n+7}{2n^3+n^2}
\cdot\frac{ \frac{1}{n^3} }{\frac{1}{n^3} } =\frac{1-
\frac{1}{n^2} +\frac{7}{n^3} }{2+\frac{1}{n} }. \] As
$n\rightarrow\infty$, both $\frac{1}{n^2} \rightarrow 0$ and
$\frac{1}{n} \rightarrow 0$, and so $\{ a_n\}$ converges to
$\frac{1}{2}$.

\item {\bf converges:} by a previous part of this exercise, we know
that $\{ (\frac{9}{10} )^n\}$ converges to $0$, since $|
\frac{9}{10} | <1$, and so $\lim_{n\rightarrow\infty} ( 1 +(
\frac{9}{10} )^n) =1+\lim_{n\rightarrow\infty} ( \frac{9}{10} )^n
=1$.

\item {\bf converges:} by a previous part of this exercise, we know
that $\{ (- \frac{1}{2} )^n\}$ converges to $0$, since $|-
\frac{1}{2} | <1$, and so $\lim_{n\rightarrow\infty} (2-(-
\frac{1}{2} )^n) =2-\lim_{n\rightarrow\infty} (- \frac{1}{2} )^n
=2$.

\item {\bf diverges:} for $n$ even, $a_n =2$, while for $n$ odd, $a_n
=0$.  In particular, $|a_n -a_{n+1}| =2$ for all $n$, and so the
sequence fails the Cauchy criterion and hence diverges.

\item {\bf converges:} note that $0\le 1+(-1)^n\le 2$ for all $n$, and
so the squeeze law yields that since $\lim_{n\rightarrow\infty}
\frac{2}{n} =0$, we have that $\lim_{n\rightarrow\infty} a_n =0$.

\item {\bf converges:} we begin by noting that
\[ 0\le \frac{ 1+(-1)^n\sqrt{n}}{ (\frac{3}{2} )^n} \le
\frac{2\sqrt{n}}{ ( \frac{3}{2})^n}, \] and so we'll concentrate
on evaluating $\lim_{n\rightarrow\infty} \frac{2\sqrt{n}}{
(\frac{3}{2})^n}$ and hope to be able to apply the squeeze law.
Since $\lim_{n\rightarrow\infty} \frac{2\sqrt{n}}{ (\frac{3}{2}
)^n}$ has the indeterminate form $\frac{\infty}{\infty}$, we may
use l'Hopital's rule to evaluate
\[ \lim_{n\rightarrow\infty} \frac{2\sqrt{n}}{( \frac{3}{2} )^n} =
\lim_{n\rightarrow\infty} \frac{ \frac{1}{\sqrt{n}}
}{\ln(\frac{3}{2}) \exp(n\ln(\frac{3}{2} ))} =
\lim_{n\rightarrow\infty} \frac{1}{\ln(\frac{3}{2} ) \sqrt{n}
(\frac{3}{2} )^n} =0 \] (where we differentiate $(\frac{3}{2} )^n$
by first writing it as $\exp(n\ln(\frac{3}{2}))$).  Hence, we may
use the squeeze law to see that $\{ a_n\}$ converges to $0$.

\item {\bf converges:} since $0\le \sin^2(n)\le 1$ for all $n$ and
since $\frac{1}{\sqrt{n}} \rightarrow 0$ as $n\rightarrow\infty$
(since $\sqrt{n}\rightarrow\infty$ as $n\rightarrow\infty$), the
comparison test yields that $\frac{\sin^2(n)}{\sqrt{n}}
\rightarrow 0$ as $n\rightarrow\infty$.  That is, $\{ a_n\}$
converges to $0$.

\item {\bf converges:} since $1\le \sqrt{2+\cos(n)}\le \sqrt{3}$ for
all $n$ and since $\frac{1}{n} \rightarrow 0$ as
$n\rightarrow\infty$, the squeeze law yields that
$\sqrt{\frac{2+\cos(n)}{n} }\rightarrow 0$ as
$n\rightarrow\infty$.  That is, $\{ a_n\}$ converges to $0$.

\item {\bf converges:} since $\sin(\pi n) =0$ for all integers $n$,
this sequence is the constant sequence $a_n =n\cdot 0 =0$ for all
$n$.  In particular, $\{ a_n\}$ converges to $0$.

\item {\bf diverges:} since $\cos(\pi n) =(-1)^n$, this sequence can
be rewritten as $a_n =(-1)^n n$.  For $n\ge 1$, $|a_{n+1} -a_n|\ge
2$, and so the sequences fails the Cauchy criterion, and so
diverges.

\item {\bf converges:} since $-1\le -\sin(n)\le 1$ for all $n$, we have
that $-\frac{1}{n} \le -\frac{\sin(n)}{n} \le \frac{1}{n}$ for all
$n$, and so $\{ -\frac{\sin(n)}{n} \}$ converges to $0$.  Hence,
$\{ a_n\}$ converges to $\pi^0 =1$.

\item {\bf diverges:} for $n$ even, $\cos(\pi n) =1$ and for $n$ odd,
$\cos(\pi n) =-1$.  In particular, $|a_{n+1} -a_n| =|2^1 -2^{-1}|
=\frac{3}{2}$ for all $n$, and so this sequences fails the Cauchy
criterion, and hence $\{ a_n\}$ diverges.

\item {\bf converges:} we could use l'Hopital's rule, since
$\lim_{n\rightarrow\infty} \frac{\ln(2n)}{\ln(3n)}$ has the
indeterminate form $\frac{\infty}{\infty}$, but we proceed in a
more low tech way.  Use the laws of logarithms and a variant of
the standard trick for rational functions, we rewrite
\[ a_n =\frac{\ln(2n)}{\ln(3n)} =\frac{\ln(2) +\ln(n)}{\ln(3) +\ln(n)}
=\frac{\ln(2) +\ln(n)}{\ln(3) +\ln(n)}\cdot \frac{\frac{1}{\ln(n)}
}{ \frac{1}{\ln(n)} } =\frac{1 + \frac{\ln(2)}{\ln(n)} }{1 +
\frac{\ln(3)}{\ln(n)} }. \] Since $\ln(n)\rightarrow\infty$ as
$n\rightarrow\infty$, we have that both $\frac{\ln(2)}{\ln(n)}$
and $\frac{\ln(3)}{\ln(n)}$ go to $0$ as $n\rightarrow\infty$, and
so $\lim_{n\rightarrow\infty} a_n =1$.

\item {\bf converges:} since $\lim_{n\rightarrow\infty}
\frac{\ln^2(n)}{n}$ has the indeterminate form
$\frac{\infty}{\infty}$, we can use l'Hopital's rule:
\[ \lim_{n\rightarrow\infty} \frac{\ln^2(n)}{n}
=\lim_{n\rightarrow\infty} \frac{2 \ln(n) \frac{1}{n} }{1}
=\lim_{n\rightarrow\infty} \frac{2 \ln(n)}{n}. \] This limit still
has the indeterminate form $\frac{\infty}{\infty}$, and we can
apply l'Hopital's rule again to get
\[ \lim_{n\rightarrow\infty} \frac{2 \ln(n)}{n}
=\lim_{n\rightarrow\infty} \frac{ \frac{2}{n} }{1} =0. \] Hence,
$\{ a_n\}$ converges to $0$.

\item {\bf converges:} write
\[ a_n =n\sin \left( \frac{1}{n} \right) = \frac{\sin( \frac{1}{n} )
}{ \frac{1}{n} }. \] Since $\lim_{n\rightarrow\infty} a_n$ has the
indeterminate form $\frac{0}{0}$, we can apply l'Hopital's rule to
get
\[ \lim_{n\rightarrow\infty} \frac{\sin \left( \frac{1}{n} \right)}{
\frac{1}{n} } =\lim_{n\rightarrow\infty} \frac{\cos \left(
\frac{1}{n} \right) \left(-\frac{1}{n^2} \right) }{-\frac{1}{n^2}
} =\lim_{n\rightarrow\infty} \cos \left( \frac{1}{n} \right)
=\cos(0) =1. \] Hence, $\{ a_n\}$ converges to $1$.  (There is
also a geometric argument for evaluating this limit, that can be
found in Adams (p. 116, Theorem 7).)

\item {\bf converges:} as $n\rightarrow\infty$, $\arctan(n)\rightarrow
\frac{\pi}{2}$, and so $\lim_{n\rightarrow\infty}
\frac{\arctan(n)}{n} =0$.  (This is an application of the squeeze
law, since the numerator is bounded by $0$ and $\pi$.)

\item {\bf converges:} since $\lim_{n\rightarrow\infty}
\frac{n^3}{e^{n/10}}$ has the indeterminate form
$\frac{\infty}{\infty}$, we may use l'Hopital's rule:
\[ \lim_{n\rightarrow\infty} \frac{n^3}{e^{n/10}} =
\lim_{n\rightarrow\infty} \frac{3n^2}{\frac{1}{10} e^{n/10}}. \]
Since this latter limit still has the indeterminate form
$\frac{\infty}{\infty}$, we use l'Hopital's rule again:
\[ \lim_{n\rightarrow\infty} \frac{3n^2}{\frac{1}{10} e^{n/10}}
=\lim_{n\rightarrow\infty} \frac{6 n}{\frac{1}{100} e^{n/10}}. \]
And as we still have the indeterminate form
$\frac{\infty}{\infty}$, we apply l'Hopital's rule yet again:
\[ \lim_{n\rightarrow\infty} \frac{6 n}{\frac{1}{100} e^{n/10}} =
\lim_{n\rightarrow\infty} \frac{6}{\frac{1}{1000} e^{n/10}}. \]
The right hand limit evaluates to $0$, and so $\{ a_n\}$ converges
to $0$.

\item {\bf converges:} write
\[ a_n =\frac{2^n+1}{e^n} = \frac{2^n}{e^n} + \frac{1}{e^n}
=\frac{2^n}{e^n} + \frac{1^n}{e^n} = \left(\frac{2}{e}\right)^n +
\left(\frac{1}{e}\right)^n. \] Since both $\frac{2}{e} <1$ and
$\frac{1}{e} <1$, we have that both $( \frac{2}{e} )^n$ and $(
\frac{1}{e} )^n$ go to $0$ as $n\rightarrow\infty$, and so their
sum goes to $0$ as $n\rightarrow\infty$.  That is, $\{ a_n\}$
converges to $0$.

\item {\bf converges:} again there are several possible approaches,
including l'Hopital's rule, but again we take a low tech approach,
and begin by expressing $\sinh(n)$ and $\cosh(n)$ in terms of
$e^n$ and $e^{-n}$, to get
\[ a_n =\frac{\sinh(n)}{\cosh(n)} =\frac{e^n -e^{-n}}{e^n +e^{-n}}
=\frac{e^n -e^{-n}}{e^n +e^{-n}}\cdot \frac{e^{-n}}{e^{-n}}
=\frac{1 -e^{-2n}}{1 +e^{-2n}}. \] Since $e^{-2n} =( \frac{1}{e^2}
)^n \rightarrow 0$ as $n\rightarrow\infty$, we see that
$\lim_{n\rightarrow\infty} a_n =1$. That is, $\{ a_n\}$ converges
to $1$.

\item {\bf converges:} as with all limits in which the variable
appears in both the base and the exponent, we begin by rewriting
using the identity $m =\exp(\ln(m))$ to get $a_n =(2n+5)^{1/n}
=\exp \left( \frac{\ln(2n+5)}{n} \right)$.  We may now use
l'Hopital's rule to evaluate the limit of the exponent
$\lim_{n\rightarrow\infty} \frac{\ln(2n+5)}{n}$ (as it has the
indeterminate form $\frac{\infty}{\infty}$) to get
\[ \lim_{n\rightarrow\infty} \frac{\ln(2n+5)}{n}
=\lim_{n\rightarrow\infty} \frac{ \frac{2}{2n+5}}{1} =0. \]
Therefore, $\{ a_n\}$ converges to $e^0 =1$.
\item {\bf converges:} as with all limits in which the variable
appears in both the base and the exponent, we begin by rewriting
using the identity $m =\exp(\ln(m))$ to get
\[ a_n = \left(\frac{n-1}{n+1}\right)^n = \left(
\frac{n+1-2}{n+1}\right)^n =\left( 1-\frac{2}{n+1} \right)^n
=\exp\left( n\ln\left( 1-\frac{2}{n+1} \right) \right). \] Since
the exponent has the indeterminate form $0\cdot \infty$ as
$n\rightarrow\infty$, we rewrite it as
\[ n\ln\left( 1- \frac{2}{n+1} \right) =
\frac{ \ln(1- \frac{2}{n+1} )}{\frac{1}{n} }, \] which as the
indeterminate form $\frac{0}{0}$ as $n\rightarrow\infty$.  We now
apply l'Hopital's rule to evaluate
\[ \lim_{n\rightarrow\infty}
\frac{ \ln \left(1-\frac{2}{n+1}\right)}{\frac{1}{n}}
=\lim_{n\rightarrow\infty} \frac{\frac{1}{1-\frac{2}{n+1}}\cdot
\frac{2}{(n+1)^2}}{-\frac{1}{n^2}} =\lim_{n\rightarrow\infty}
\frac{-2n^2}{\left( 1-\frac{2}{n+1} \right)\cdot (n+1)^2} = -2. \]
Hence, $\{ a_n\}$ converges to $e^{-2}$.

\item {\bf converges:} since $-\frac{1}{n} \rightarrow 0$ as
$n\rightarrow\infty$, we see that $\{ a_n\}$ converges to
$(0.001)^0 =1$.

\item {\bf converges:} as $n\rightarrow\infty$, $\frac{n+1}{n} =1+
\frac{1}{n} \rightarrow 1$, and so $\{ a_n\}$ converges to $2^1
=2$.

\item {\bf converges:} one way to evaluate this limit is to write $a_n
=( \frac{2}{n} )^{3/n} = \frac{2^{3/n}}{n^{3/n}}$ and to evaluate
the limits of the numerator and denominator separately.  To
evaluate $\lim_{n\rightarrow\infty} 2^{3/n}$, all we need note is
that $\lim_{n\rightarrow\infty} \frac{3}{n} = 0$, and so $\{
2^{3/n}\}$ converges to $2^0 =1$.

\medskip
\noindent To evaluate $\lim_{n\rightarrow\infty} n^{3/n}$, we
rewrite $n^{3/n}$ as $n^{3/n} =\exp(\ln(n) \frac{3}{n})$ and use
l'Hopital's rule to evaluate $\lim_{n\rightarrow\infty} \frac{3
\ln(n)}{n}$ (since it has the indeterminate form
$\frac{\infty}{\infty}$).  Using l'Hopital's rule, we get that
\[ \lim_{n\rightarrow\infty} \frac{3\ln(n)}{n}
=\lim_{n\rightarrow\infty} \frac{ \frac{3}{n} }{1} =0, \] and so
$\{ n^{3/n}\}$ converges to $e^0 =1$.  Therefore,
\[ \lim_{n\rightarrow\infty} \frac{2^{3/n}}{n^{3/n}} =
\frac{\lim_{n\rightarrow\infty} 2^{3/n}}{\lim_{n\rightarrow\infty}
n^{3/n}} =\frac{1}{1} =1. \]

\item {\bf diverges:} begin by ignoring the $(-1)^n$ and worrying
about what happens to the rest of the term.  Using the standard
trick, massage to get $(n^2+1)^{1/n} =\exp( \frac{\ln(n^2 +1)}{n}
)$.  Since $\lim_{n\rightarrow\infty} \frac{\ln(n^2 +1)}{n}$ has
the indeterminate form $\frac{\infty}{\infty}$, we may use
l'Hopital's rule to evaluate
\[ \lim_{n\rightarrow\infty} \frac{\ln(n^2 +1)}{n}
=\lim_{n\rightarrow\infty} \frac{ \frac{2n}{n^2 +1} }{1} =0, \]
and so
\[ \lim_{n\rightarrow\infty} \exp \left( \frac{\ln(n^2 +1)}{n} \right)
=e^0 =1. \] So, putting the $(-1)^n$ back into the picture, we see
that $\{ a_n\}$ fails the Cauchy criterion: specifically, since
$\{ \frac{n^2+1}{n} \}$   converges to $1$, for any $\varepsilon
>0$, there exists $M$ so that $\left| \frac{n^2+1}{n} -1 \right|
<\varepsilon$ for $n>M$. Choose $\varepsilon =\frac{1}{2}$, and
note that for $n>M$, we get that $|a_n -a_{n+1}| > 1$, since one
of $a_n$, $a_{n+1}$ is within $\frac{1}{2}$ of $1$ and the other
is within $\frac{1}{2}$ of $-1$ (remember the alternating signs).
So, $\{ a_n\}$ diverges.

\item {\bf converges:} we perform a bit of algebraic massage: note
that
\[ a_n =\frac{ \left( \frac{2}{3} \right)^n}{ \left( \frac{1}{2}
\right)^n+ \left( \frac{9}{10} \right)^n} < \frac{ \left(
\frac{2}{3} \right)^n}{ \left( \frac{9}{10} \right)^n}
=\left(\frac{20}{27} \right)^n. \] Since $\left(\frac{20}{27}
\right)^n\rightarrow 0$ as $n\rightarrow\infty$ (since
$\frac{20}{27} <1$), the comparison test yields that $\{ a_n\}$
converges to $0$ as well.
\end{enumerate}


\end{document}
