-1

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

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.

Xetrov
  • 2,089
Rut
  • 11
  • Hint: Root Test... – sranthrop Apr 05 '18 at 15:55
  • @Rut Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 07 '18 at 20:06

2 Answers2

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.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
1

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

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

user
  • 154,566