# True or false: There exists a series $\sum_n a_n$ of non-negative terms that is convergent, such that $\sum (a_n)^{5/6}$ diverges.

So far I'm thinking that we can use a p-series, $$\sum \frac{1}{n^p},$$ which converges if and only if $p>1$. How would I then show the series where each term is raised to the power of $5/6$ diverges? Thanks.

You can use that $p$-series for any $p \in \mathbb{R}$ such that

$$1 < p \leq 1.2$$

The resulting $p$-series will have terms

$$\frac{1}{n} >a_n \geq \frac{1}{n^{1.2}}$$

And the series with terms raised to the $5/6^{\text{th}}$ power, will have

$$\left(a_n\right)^{5/6}\geq\frac{1}{n}$$

The former converges because it is a $p$-series with $p>1$. The latter diverges because it is a $p$-series with $p\leq 1$. You can also see that the second series diverges because its terms are all at least as big as the terms of the Harmonic series, and that series diverges.

$\sum \dfrac{1}{n^{6/5}}$ converges. But $\sum \left(\dfrac{1}{n^{6/5}}\right)^{5/6}=\sum \dfrac{1}{n}$ diverges.