Convergence of $\sum_{n=1}^{\infty }\left ( 1+\frac{1}{n} \right )^{n^{2}}\cdot \frac{(-1)^n}{ne^{n}}$ Why is $\sum_{n=1}^{\infty }\left ( 1+\frac{1}{n} \right )^{n^{2}}\cdot \frac{(-1)^n}{ne^{n}}$ convergent and $\sum_{n=1}^{\infty }\left ( 1+\frac{1}{n} \right )^{n^{2}}\cdot \frac{1}{ne^{n}}$ divergent? Which convergence tests can I use to show that?
Thanks in advance.
 A: We have
\begin{align}u_n=\left ( 1+\frac{1}{n} \right )^{n^{2}} \frac{(-1)^n}{ne^{n}}&=\frac{(-1)^n}{ne^n}e^{n^2\log(1+\frac{1}{n})}=\frac{(-1)^n}{ne^n}e^{n^2(\frac{1}{n}-\frac{1}{2n^2}+o(\frac{1}{n^2}))}\\&=\frac{(-1)^n}{n}-\frac{(-1)^n}{2n^2}+o(\frac{1}{n^2})\end{align}
hence the series $\displaystyle \sum_{n=1}^\infty u_n$ is convergent as sum of convergent series, moreover since 
$$|u_n|\sim_\infty \frac{1}{n}$$
then the series $\displaystyle \sum_{n=1}^\infty |u_n|$ is divergent.
A: We have
$$
\lim_{n\to\infty}\frac{1}{e}\Bigl(1+\frac1n\Bigr)^n=1.
$$
Now
$$
\log\Bigl(\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}\Bigr)=n^2\log\Bigl(1+\frac1n\Bigr)-n.
$$
Using the Taylor expansion of $\log(1/(1+x))$ we get
$$
\log\Bigl(1+\frac1n\Bigr)\sim\frac1n-\frac1{2\,n^2}.
$$
This proves
$$
\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}=e^{-1/2}=\frac{1}{\sqrt e}.
$$
By the comparison criterion, the second series is divergent.
The first series is alternate. Clearly
$$
\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}\frac1n=0.
$$
To prove convergence, by Leibniz's criterion, it is enough to show that
$$
\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}\frac1n
$$
is decreasing. I am sure it is true, but I have not checked the details.
