# Does $\sum_{n=1}^\infty \frac{(-1)^n}{n^{1+\frac{1}{n}}}$ converge?

I want to use the alternating series test here, but I've just been told that it won't work because it's not monotonically decreasing.

However, if the alternating harmonic series converges then don't we have for $\sum_{n=1}^\infty \frac{(-1)^n}{n^{1+\frac{1}{n}}}$ that $$\lim_{n \to \infty} \frac{1}{n^{1+\frac{1}{n}}} = 0$$ since $$\lim_{n \to \infty} \frac{1}{n^{1+\frac{1}{n}}} < \lim_{n \to \infty} \frac{1}{n} = 0.$$

Can someone point out where the mistake here is?

To show that it is monotonically decreasing one should show that: $$\frac{1}{n^{1+\frac{1}{n}}} > \frac{1}{(n+1)^{1+\frac{1}{n+1}}}.$$ This is equivalent to showing that: $$\frac{n+1}{n} > \frac{n^\frac{1}{n}}{(n+1)^\frac{1}{n+1}},$$ which is the same as $$(1+\frac{1}{n})^n > \frac{n}{n+1}\cdot (n+1)^\frac{1}{n+1}.$$
For sufficiently large values of $n$, this must be the case, as the limit of the LHS is just $e$ and the one of the RHS is $1$.
I’m not entirely sure what you’re proposing as an argument, but the following is not a theorem (and it looks like you may think it is): If $0\le a_n \le b_n$ and $\displaystyle\sum_{n=1}^\infty (-1)^n b_n$ converges, then $\displaystyle\sum_{n=1}^\infty (-1)^n a_n$ converges. For a counterexample, let $b_n=\frac{1}{n}$ and let $a_n$ be $0$ for even $n$ and $\frac{1}{n}$ for odd $n$.
That said, the function $f(x)=x^{(1+\frac{1}{x})}$ is increasing* for $x>0$, so the terms of your sequence decrease in absolute value and the alternating series test hypotheses are true.
*The function $f(x)$ is differentiable for $x>0$, and for positive $x$, its derivative, $\displaystyle x^{1 + \frac{1}{x}} \left(\frac{1 + \frac{1}{x}}{x} - \frac{\log x}{x^2}\right)$, is greater than $x(\frac{1}{x}-\frac{1}{x})$, and therefore positive.
I haven't checked whether or not $\frac{1}{n^{1+\frac{1}{n}}}$ is a monotonically decreasing sequence, but I will point out that $\displaystyle\lim_{n\to \infty} \frac{1}{n^{1+\frac{1}{n}}} = 0$ does not imply that $\frac{1}{n^{1+\frac{1}{n}}}$ monotonically decreases. You must confirm that both properties hold. By the way, it would be enough to know that $\frac{1}{n^{1+\frac{1}{n}}}$ monotonically decreases after some point (i.e. for all $n > M$ for some fixed $M$).