Finding convergence for an infinite sum

Question

$$\sum_{k=1}^\infty(\sqrt[k] k - 1)^{2k}$$

I have to find convergence using the tests I know. (divergence,integral,ratio,root,comparison,limit comparison) My issue is I can't figure out how to not get an inconclusive test result.

Answer 1

What about the $\;k\,-$ th root test?:

$$\sqrt[k]{\left(\sqrt[k]k-1\right)^{2k}}=\left(\sqrt[k]k-1\right)^2\xrightarrow[k\to\infty]{}0$$

so the series converges.

Answer 2

Note that since $\frac{\log k}{k}\to 0$

$$(\sqrt[k] k - 1)^{2k}=\left(e^{ \frac{\log k}{k}}-1\right)^{2k}\sim \left(1+{\frac{\log k}{k}-1}\right)^{2k}=\left({\frac{\log k}{k}}\right)^{2k}\to 0$$