Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

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

share|cite|improve this answer
Maybe ${a_n}\geqslant{0}$ ? – M. Strochyk Jan 9 '13 at 14:47
Do we know that the $a_n$ are nonnegative? Otherwise something like $a_n=(-1)^n/n$ can cause problems... – Clayton Jan 9 '13 at 14:48
because he need $\sqrt[4]{*}$,so i think it implies $a_n \geq 0$ – Laura Jan 9 '13 at 14:48
@Tai: Since it is a question of convergence, not a question of showing the series converges, I don't think we can assume $a_n\geq0$. – Clayton Jan 9 '13 at 14:51
Supposing of course non-negative terms, the last inequality isn't true in general, while the conclusioin remains true. Take for example, $a_n=(100,\frac{1}{4},\frac{1}{9},...\frac{1}{n^2}...)$. – sheriff Jan 9 '13 at 15:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.