Absolute convergence and conditional convergence of the sum $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\cdot(1+\frac{1}{n})^n}{n}$ I am trying to solve an exercise from an old exam from my university which no solution is uploaded.
Basically, I have to determine if the series:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\cdot(1+\frac{1}{n})^n}{n}$$
is convergent and also check if it is absolutely convergent.
To know if it is convergent I am trying to apply the Leibnitz Criteria for alternating series, that states that:
$\sum_{n=1}^{\infty} (-1)^{n}a_n$ convergent iff $a_n$ is decreasing and its limit is $0$
For this series, if we call $a_n = \frac{(1+\frac{1}{n})^n}{n}$ I have proven that it converges to $0$ but I do not see an easy way to prove that It is decreasing.
For the absolute convergence I have tried several criteria but none of them works for me.
Thank you in advance.

Yes, the sequence $a_n=\frac{\left(1+\frac1n\right)^n}{n}$ is decreasing and the proof is fairly simple:
$$\frac{a_{n+1}}{a_n} =\frac{(\frac{n+2}{n+1})^{n+1}}{n+1}\cdot\frac{n}{(\frac{n+1}{n})^n} =\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}<1$$
where at the last step we just note that $n(n+2)=(n+1)^2-1<(n+1)^2$.
Since you already know that $a_n\to 0$, by Leibniz criterion, the series is convergent.
The series is NOT absolute convergent by limit comparison test:
$$\left|\frac{(-1)^{n+1}\cdot(1+\frac{1}{n})^n}{n}\right|\sim \frac{e}{n}$$
and $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent.
