[Math] Does the series $\sum_{k=1}^{\infty}\left(\frac{1}{k^{\log k}}\right)$ converge

Question

$$\sum_{k=1}^{\infty}\left(\frac{1}{k^{\log k}}\right)$$ this is converge. But how? I think here we use comparison thm.

Answer 1

Try comparing this series to $$\sum_{k=1}^\infty \frac{1}{k^2}.$$ Since $\log(x)\geq 2$ for $x\geq e^2$, after throwing out finitely many terms you can prove convergence.

Added: Alternatively, by using the Cauchy condensation test, we need only prove the convergence of $$\sum_{n=0}^{\infty}\frac{2^n}{2^{n^{2}\log2}}.$$