Does $\sum\limits_{k=1}^∞\frac1{k^{\ln k}}$ converge or diverge?

Question

$$\sum_{k=1}^{\infty} \frac{1}{k^{\ln k}}.$$

I have some problems with this sum, I have tried most of the common methods, such as the ratio/root test, integral test, etc. I do believe it looks like a problem easily solvable by a direct comparison with some p-series, however, I do not know how to formulate the inequality.

I did try using the common fact that $\ln(x) \ll x^p$ for large $x$ but that of course only bounds it from below with a convergent series saying nothing about ours. Also I tried rewriting the 'summand' as $e^{\ln(k^{\ln k})}=e^{\ln k \cdot \ln k}$ but this did not really yield anything.

Any answer or hint would be greatly appreciated.

Answer 1

HINT: Notice that for $k\ge 3$, it is true that $\ln(k)\gt 1$.

Answer 2

Hint: By the Cauchy-condensation test, the given series converges if and only if the following series converges $$ \sum_{n=0}^\infty 2^n\frac{1}{(2^{n})^{n\ln 2}}= \sum_{n=0}^\infty \frac{2^n}{2^{n^2\ln2}} $$

Answer 3

For $k > e^2$ we have $\ln(k) > 2 $ and $ k^{\ln(k)} > k^2 $

$$ \frac{1}{k^{\ln(k)}} < \frac {1}{k^2} \quad \rightarrow \sum_k \,\frac{1}{k^{\ln(k)}} < \infty $$