Does this series converge or not?

Question

$$\sum _{k=2}^{\infty }\:\frac{1}{\sqrt{k}\left(\ln k\right)^{\ln k}}$$

I've tried limit comparison with $\frac{1}{\sqrt{n}}$ and $\frac{1}{n^2}$ and also Cauchy condensation test but nothing seems to work. With the condensation test I reached this point $$\frac{2^k}{k\cdot \sqrt{2^k}\cdot c^k}$$ where $c=\left(\ln2\right)^{\ln2}$ and then got stuck. Root and fracion tests create issues with the ln so I can't think of anything else

Answer 1

For each $k\in\Bbb N$,$$\frac{2^k}{k\sqrt{2^k}c^k}=\frac1k\left(\frac{\sqrt2}c\right)^k\to\infty,$$since $\frac{\sqrt2}c>1$. Therefore, your series diverges.

Answer 2

Note that for $k > e^{e^2}$ we have $\ln \ln k > 2$ and

$$(\ln k )^{\ln k} = e^{\ln \ln k \cdot \ln k}= k ^{\ln\ln k}> k^2$$

Thus, for all sufficiently large $k$

$$\frac{1}{\sqrt{k}(\ln k )^{\ln k}} < \frac{1}{\sqrt{k} k^2}< \frac{1}{k^2},$$

and the series converges by the comparison test.