Does the series $\sum_{n\ge1}\frac{(-1)^n}{\left(n\ln\frac{n+1}n\right)^n}$ converge? It seems to me that the general term of this series is not tends to zero :
$\left(n\ln\left(1+\frac1n\right)\right)^n\sim n^n\frac1{n^n}=1$
so
$\frac 1{\left(n\ln\frac{n+1}n\right)^n}\ge1$.
Am I right ?
Thanks.
 A: The series fails to converge because the general term does not tend to zero. Let $a_n = \dfrac{(-1)^n}{\left(n\ln(1+\frac{1}{n})\right)^n}$. Consider the sub sequence $(a_n)$ with with $n$ even. For the case $n$ is even, $(-1)^n = 1$, so $a_n = \dfrac{1}{\left(n\ln(1+\frac{1}{n})\right)^n} = \dfrac{1}{\ln\left((1+\frac{1}{n})^n\right)^n}$. But $1 < (1+\frac{1}{n})^n < e$ $\forall n$ ( can be proved by induction on $n$). So: $\ln(1+\frac{1}{n})^n < \ln e = 1$, this implies that $\dfrac{1}{\ln(1+\frac{1}{n})^n} > 1$ $\forall n$, so $a_{2n} > 1$ $\forall n$ . So: $\displaystyle \lim_{n\to \infty} a_{2n} \geq 1$. Since there is a subsequence of $(a_n)$ that does not converge to $0$, the sequence itself $(a_n)$ cannot converge to $0$, which means the series cannot converge.
A: There are several problems with the proposed attempt:


*

*if $a_n\sim b_n$, it's not necessarily true that $a_n^n\sum b_n^n$ (for example take $a_n=\frac 1n$ and $b_n=\frac 1{n+1}$);

*if $a_n\sim b_n$ and $a_n,b_n\gt 0$, it's not necessarily true that $a_n\geqslant b_n$ for $n$ large enough (but it is true that $a_n\geqslant b_n/2$ for $n$ large enough). 


However, using the inequality $\log (1+x)\leqslant x$ valid for $x\geqslant 0$ with $x:=1/n$, we obtain $n\log (1+1/n)\leqslant 1$, hence 
$$\left(n\log\left(1+\frac 1n\right) \right)^n\leqslant 1$$
and the general term of the series has an absolute value greater or equal to $1$, hence the series fails to converge.
