# If one series converge, the other also converge?

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?

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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.

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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

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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