Question: If $\sum\_{n=1}^{\infty}a\_n$ converge, then

$\sum\_{n=1}^{\infty}\sqrt[4]{a\_{n}^{5}}=\underset{n=1}{\overset{\infty}{\sum}}a\_{n}^{\frac{5}{4}}$ converge?

Please verify answer below.

Answer: 
Yes. Because $\sum a\_n$ converges, we know that the limit $\lim a\_n = 0$, by the Divergence Test.

This means that $\exists N, \forall n > N, a\_n <1 $. It stands to reason then that $a\_n^{5/4} \le a\_n$ for all such $n>N$. By Direct Comparison, then, $\sum a\_n^{5/4}$ also converges.

We cannot say $\sum a\_n^{5/4} < \sum a\_n$. Someone said that but it is untrue. It all depends on the front-end behavior. If for the first $N$ terms $a\_n^{5/4}$ is sufficiently larger than $a\_n$ then the inequality of the actual sum may be reversed.

---

It doesnt matter if for any terms $a\_n<0$. If a term is odd then $a\_n^5$ is still odd. And the principle fourth root $a\_n^{5/4}$ is going to be a complex number where $a\_n^{5/4} = |a\_n|^{5/4} \frac{\sqrt{2}+i\sqrt{2}}{2}$ for all negative $a\_n$

The principle fourth root $a\_n^{5/4}$ for positive $a\_n$ is still positive and real.

Simply break the series apart into two separate series of positive and negative $a\_n$ terms. We know that $\sum\_{a\_n>0} a\_n^{5/4}$ is going to converge as per the first part of this proof. It has fewer terms and therefore converges to a smaller sum.

For the series $\sum\_{a\_n<0} a\_n^{5/4}$ we get the final sum $\frac{\sqrt{2}+i\sqrt{2}}{2} \sum\_{a\_n<0} |a\_n|^{5/4} $. Notice the sum is still going to be convergent as per the first part of this proof, but there is a complex scalar that was factored out.

Thus, even for a series containing negative $a\_n$ terms, the sum $\sum a\_n^{5/4}$ is simply going to be the sum $\left(\sum\_{a\_n>0} a\_n^{5/4}\right) + \left(\frac{\sqrt{2}+i\sqrt{2}}{2} \sum\_{a\_n<0} |a\_n|^{5/4}\right) $

Each individual series is now a series of positive terms, each containing fewer terms than the original series and therefore each converges to a smaller sum.

---

**Disclaimer** Im no expert in infinite series. They are not exactly intuitive. 

The problem is infinitely more complicated if you let $a\_n$ take any complex value.
Because $\sum\_{n=1}^{\infty}a\_{n}$ converge, then $a\_n\rightarrow0$, so for big $n$: $a\_{n}<1$, which implies $$\sum\_{n=1}^{\infty}a\_{n}^{\frac{5}{4}}<\sum\_{n=0}^{\infty}a\_{n}$$ so it's also converge