# How can I prove that $\sum_{k=1}^{\infty}\frac{(2\sqrt[k]k-1)^k}{k^4}$ is convergent?

I'm trying to prove that the sum $\sum_{k=1}^{\infty}\frac{(2\sqrt[k]k-1)^k}{k^4}$ is convergent.

I've tried Cauchy's root test - but I get the limit to be 1, so the test is inconclusive. I also tried using d'Alemberts ratio test, but that didn't provide any information either. I can't use the integral test, because I can't find the antiderivative. The idea I had was to compare it to $\sum_{k=1}^{\infty} \frac{1}{k^2}$. I know this series converges and I'm pretty confident that $\frac{(2\sqrt[k]k-1)^k}{k^4}\le \frac{1}{k^2}$ for all $k \ge 1$ but I don't know how to prove that.

I would be really grateful for any ideas!

Notice $$\sqrt[k]{k^2} - (2\sqrt[k]{k} - 1) = (\sqrt[k]{k} - 1)^2 \ge 0 \implies \sqrt[k]{k^2} \ge 2\sqrt[k]{k} - 1\tag{*1}$$ This leads to $$\sum_{k=1}^N\frac{\left(2\sqrt[k]{k}-1\right)^k}{k^4} \le \sum_{k=1}^N\frac{\left(\sqrt[k]{k^2}\right)^k}{k^4} = \sum_{k=1}^N\frac{1}{k^2} \le \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$ This means the partial sums are bounded from above. Since the terms are non-negative, the series converges to some number no more than $\displaystyle\;\frac{\pi^2}{6}$.

We have: $$k^{1/k} = e^{\frac{\log k}{k}} \leq \frac{1}{1-\frac{\log k}{k}}$$ hence: $$2k^{1/k}-1 \leq \frac{1+\frac{\log k}{k}}{1-\frac{\log k}{k}}$$ and for any $k$ big enough: $$2k^{1/k}-1 \leq \exp\left(\frac{5}{2}\frac{\log k}{k}\right)$$ so: $$(2k^{1/k}-1)^k \leq k^{5/2}$$ ensuring convergence.

• Thanks for your answer! Can you explain the step from the second to the third inequality? – Jarvi79 Sep 21 '14 at 19:18