Testing series for convergence/divergence The task is to test the following series for convergence/ divergence:
$$\sum_{n=1}^\infty \frac{(a+nx)^n}{n!}$$
Now, I have been able to use the Ratio Test and establish that the series converges for $x<1/e$ and diverges for $x>1/e$, but testing the series at $x=1/e$ has been a little more challenging. Could someone tell me how I might get the job done?  
 A: Let $x=\frac{1}{e}$.
Since $$
\frac{(a+\frac{n}{e})^n/n!}{(\frac{n}{e})^n/n!}=\left(1+\frac{ae}{n}\right)^n\to e^{ae},
$$ by the comparison test, $\sum_{n=1}^\infty a_n<\infty $ if and only if $\sum_{n=1}^\infty \frac{(\frac{n}{e})^n}{n!}<\infty$. So we may assume that $a=0$. By Stirling's formula, we have $$\lim_{n\to\infty}\frac{n!}{\sqrt{2\pi n}(\frac{n}{e})^n}=1.$$ Since it is a positive sequence with a positive limit, the sequence should be bounded away from $0$ (i.e. have a positive infimum) and have a bounded supremum. So there exist $c>0$ and $C>0$ such that
$$c\le \frac{\sqrt{n}(\frac{n}{e})^n}{n!}\le C,
$$or equivalently
$$
\frac{c}{\sqrt{n}}\le \frac{(\frac{n}{e})^n}{n!}\le \frac{C}{\sqrt{n}}.
$$ 
Since $\sum_n \frac{1}{\sqrt{n}}=\infty$, the series diverges for $x=\frac{1}{e}$.
If $x=-\frac{1}{e}$, then the series becomes alternating eventually. Therefore, the series converges if and only if $|a_n|\to 0$ as $n\to \infty$. And this follows immediately from Stirling's formula:
$$\begin{eqnarray}
\lim_{n\to\infty}|a_n|&=& \lim_{n\to\infty}\frac{(\frac{n}{e}-a)^n}{n!}\\
&=&\lim_{n\to\infty}\frac{(\frac{n}{e}-a)^n}{\sqrt{2\pi n}(\frac{n}{e})^n}\\
&=&\lim_{n\to\infty}\frac{1}{\sqrt{2\pi n}}\left(1-\frac{ae}{n}\right)^n=0.
\end{eqnarray}$$
There is an alternative approach avoiding use of Stirling's formula. Note that by Taylor series expansion, we have
$$
\log(1+t) = t-\frac{t^2}{2}+o(t^2).
$$ This implies that there exists $\delta>0$ such that
$$
\exp\left(t-ct^2\right)\le 1+t\le \exp\left(t-Ct^2\right),\quad\forall 0\le t\le \delta
$$ for some $c\in (\frac{1}{2},1)$ and $C\in (0,\frac{1}{2})$. Let $$a_n = \frac{(\frac{n}{e})^n}{n!}.$$ Then
$$
\frac{a_{n+1}}{a_n}=\frac{1}{e}\left(1+\frac{1}{n}\right)^n,
$$ and hence
$$
\exp\left(-\frac{c}{n}\right)\le\frac{a_{n+1}}{a_n}\le \exp\left(-\frac{C}{n}\right)
$$ for all sufficiently large $n$. This gives for all large $n$,
$$
k\exp\left(-c\sum_{j=1}^{n-1}\frac{1}{j}\right)\le a_n\le K\exp\left(-C\sum_{j=1}^{n-1}\frac{1}{j}\right)
$$ for some $k>0$ and $K>0$. Using the fact that $\int_j^{j+1}\frac{dt}{t}\le\frac{1}{j}=\int_{j-1}^j\frac{1}{j}dt\le\int_{j-1}^j\frac{dt}{t}$ for $j\ge 2$, we have
$$
\log n=\int_1^n\frac{dt}{t}\le\sum_{j=1}^{n-1}\frac{1}{j}\le 1+\int_1^{n-1}\frac{dt}{t}\le 1+\log n.
$$ This in turn implies
$$
ke^{-c}\frac{1}{n^c}\le a_n \le \frac{K}{n^C}.
$$ Now, if $x=\frac{1}{e}$, then $\sum_n a_n =\infty$ follows from $a_n\ge ke^{-c}\frac{1}{n^c}$ for all but finitely many $n$. If $x=-\frac{1}{e}$, then $\frac{|a-\frac{n}{e}|^n}{n!}\to 0$ follows from
$$\begin{eqnarray}
\lim_{n\to\infty}\frac{|a-\frac{n}{e}|^n}{n!}&=& \lim_{n\to\infty}\frac{(\frac{n}{e}-a)^n}{n!}\\
&=&e^{-ae}\lim_{n\to\infty}a_n\\
&\le&e^{-ae}\lim_{n\to\infty}\frac{K}{n^C}=0.
\end{eqnarray}$$
A: When $x=1/e$ the numerator is equivalent (as $n\to \infty$) to $(n/e)^n$. Using Stirling's formula for $n!$ your general term is equivalent to $1/\sqrt{2\pi n}$ and your series is therefore divergent.
