Problem statement: Does the series $$\sum^\infty_{n=1} \frac{(-1)^n}{n^{1+\frac{1}{n}}}$$ diverge, converge, or converge absolutely?
EDIT: I appreciate all the answers so far. In my book, we haven't reached the Dirichlet test, Leibniz test, or derivatives, so if anyone has a line of reasoning that does not use these techniques, I would be interested to see it.
Solution attempt: We have $$\left|\frac{(-1)^n}{n^{1+\frac{1}{n}}}\right|=\frac{1}{n^{1+\frac{1}{n}}}=\frac{1}{n^{\phantom{\frac{1}{n}}}}\frac{1}{n^{\frac{1}{n}}}$$ Let ${\{b_n\}}=\frac{1}{n^{\frac{1}{n}}}$. Then $\frac{1}{\{b_n\}}=n^{\frac{1}{n}}\to 1,$ so $\frac{1}{\{b_n\}}$ is bounded, but also $\sum{|\frac{1}{n}|}$ diverges. By Theorem 26.4(ii) in Foundations of Mathematical Analysis by Johnsonbaugh, the series does not converge absolutely.
Thus, we need to determine if the series converges conditionally or diverges. At this point, I tried applying the root test, the ratio test, and the comparison test, but I didn't see anything that was tractable ... but maybe someone else does.
To apply the alternating series test, I would need to argue that $\frac{1}{n^{1+\frac{1}{n}}}$ is decreasing with limit zero. I think that this sequence does have limit zero, but I'm not sure how to show that it is decreasing, since I would need to compare $a_{n+1}=\frac{1}{(n+1)^{1+\frac{1}{n+1}}}$. I'm not sure how to handle this because $(n+1)$ is bigger, but $1+\frac{1}{n+1}$ is a smaller exponent.
Any hints?