<s>[INST] Show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges How would you show that  $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges?

My approach
At first glance I would go for the alternate series criterion and show that $\frac{\left(1+\frac{1}{n}\right)^n}{n}$ is a monotone sequence with $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$. So after a few algebraic manipulations I get:
$$
\frac{\frac{\left(1+\frac{1}{n}\right)^n}{n}}{\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{n+1}}=
\frac{\left(1+\frac{1}{n}\right)^n(n+1)}{\left(1+\frac{1}{n+1}\right)^{n}\left(1+\frac{1}{n+1}\right)n}=
\left(\frac{1+\frac{1}{n}}{1+\frac{1}{n+1}}\right)^{n}\cdot \frac{n+1}{\frac{n^2+2n}{n+1}}=
\cdots=\frac{n^2+2n+1}{n^2+2n}\left(\frac{n^2+2n+1}{n^2+2n}\right)^n\geq1.
$$
So the expression is monotonically decreasing. Further, we know that $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e^1$. Hence, $\left(1+\frac{1}{n}\right)^n$ is bounded, say $S\geq\Big| \left(1+\frac{1}{n}\right)^n\Big|$. This allows us to conclude:
$$
\Big| \frac{\left(1+\frac{1}{n}\right)^n}{n}-0\Big|\leq \frac{S}{n}<\epsilon,\text{ where n is sufficiently large enough}.
$$
Hence, $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$ and by the alternate series crterion the series $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges.
Is this correct? Is there a faster way?
 [/INST] Since
$$
2\le\left(1+\frac1n\right)^n\le3
$$
and
$$
\sum_{n\ge1}\frac{(-1)^n}n=\log2
$$
Calling $S$ the desired limit (that is, your sum), you can immediately get that
$$
2\log 2\le S\le3\log 2\;.
$$
</s>