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{\bf Question}

Decide which of the following series is convergent
\begin{description}
\item[(a)]$\ds 1 + x +  \frac{x^2}{2 \times 1} + \frac{x^3}{3 \times 2 \times
1}+ ... + \frac{x^n}{n!} + ...$
\item[(b)]$ 100 + 10 +1 +0.1 + ... +10^{2-n} + ...$
\item[(c)]$1-2+3-4+...+(-1)^{n-1}n+...$
\item[(d)]$1+2+3+4+...+n+... $
\end{description}


\vspace{.25in}

{\bf Answer}

\begin{description}



\item[(a)]
$\ds u_n =\frac{x^n}{n!} \hspace{.4in} S_n = 1 + x + ... +
\frac{x^n}{n!}$

$\ds \frac{u_{n+1}}{u_n} = \frac{x}{n+1} < r < 1$ for sufficiently
large $n$.

Hence $\ds \lim_{ n \to \infty} S_n = 1 + x + \frac{x^2}{2!} + ...
+ \frac{x^n}{n!} + \frac{x^{n+1}}{(n + 1)!}+....$

$\ds 0 < S < 1 + x + \frac{x^2}{2!} + ... + \frac{x^n}{n!} (1 + r
+ r^2 + ...)$ for some $n$


$\ds \Rightarrow 0 < S < S_{n-1} + \frac{x^n}{n!(1-r)}$

So this is convergent (by d'Alembert's test)




\item[(b)]
By the formula  with $r = 0.1$ $u_1 = 100$

$\ds S = \frac{100}{1 - 0.1} = \frac{u_1}{1 - r} =
\frac{100}{0.9}$

so this series is convergent




\item[(c)]
$\ds S =  1 - 2 + 3 - 4 + ...  \hspace{.4in} S_n = \frac{n}{2}[1 +
(-1)^nn]$

$|u_n|$ does not tend to zero, so S cannot exist Therefore this
series is divergent.




\item[(d)]
$\ds S = 1 + 2 + 3 + 4 + .... \hspace{.3in} u_n = n$

$\ds S_n = \frac{n}{2}[1 + n] \hspace{.3in} \lim_{n \to \infty}
S_n = \infty$

So this series is divergent.

\end{description}


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