sequences and series - 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?

Answer

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 $1$.