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This is the $1988$ Putnam $B4$ Problem:

Prove that if $\sum_{n=1}^\infty a_n$ is a convergent series of positive real numbers, then so is $\sum_{n=1}^\infty a_n^{n/({n+1})}$.

My problem lies in figuring out what to do in the case that $0\lt a_n \lt 1$ for all $n$. I imagine that it must have something to do with the limit comparison test. Any hints would be greatly appreciated.

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    $\begingroup$ Note: You can find many Putnam answers, including this one, here. $\endgroup$ – vadim123 Nov 9 '14 at 21:10
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    $\begingroup$ Erm . . If each $a_n \ge 1$, then $\sum_{n=1}^\infty a_n$ diverges. . . $\endgroup$ – user88319 Nov 9 '14 at 21:11
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    $\begingroup$ Yeah, your "trivial case" is not a case at all. $\endgroup$ – Thomas Andrews Nov 9 '14 at 21:13
  • $\begingroup$ @Strants Thanks, I removed the misleading statement about the "trivial" case $\endgroup$ – Josh Gray Nov 9 '14 at 21:19
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Limit comparison seems good. What can we say about $$\lim_{n\to\infty}\frac{a_n}{a_n^{n/(n+1)}}=\lim_{n\to\infty}a_n^{1/(n+1)}?$$

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  • $\begingroup$ Does the right hand side go to zero since $a_n$ converges? $\endgroup$ – Josh Gray Nov 9 '14 at 21:22
  • $\begingroup$ You don't want it to go to zero, really you just want it to have a finite $\limsup$ - that is $\limsup a_n^{1/(n+1)}<+\infty$. $\endgroup$ – Thomas Andrews Nov 9 '14 at 21:23
  • $\begingroup$ Not zero, but can you argue it's got a finite limit (or even lim sup)? $\endgroup$ – Ted Shifrin Nov 9 '14 at 21:24
  • $\begingroup$ Sorry for reviving, but I don't see how the application works here - shouldn't our limits be the opposite way round? According to the following post: math.stackexchange.com/questions/1761541/… $\endgroup$ – Bryan Shih Jun 13 '16 at 0:36
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Let $A=\{n: a_n \le2^{-n-1}\}$ and $B=\{n:a_n>2^{-n-1}\}$.

If $n\in A$, then $a^{n/(n+1)}_n\le 2^{-n}$ and hence $$ \sum_{n\in A}a^{n/(n+1)}_n\le \sum_{n\in A}2^{-n}\le 1. $$

If $n\in B$, then $a_n> 2^{-n-1}$, then $2^{n+1}a_n> 1$, and thus $2a_n^{1/(n+1)}>1$ and therefore $2a_n>a_n^{n/(n+1)}$. Hence $$ \sum_{n\in B}a_n^{n/(n+1)}\le 2\sum_{n\in\mathbb N}a_n. $$ Thus $$ \sum_{n\in\mathbb N}a_n^{n/(n+1)}= \sum_{n\in A}a_n^{n/(n+1)}+\sum_{n\in B}a_n^{n/(n+1)} \le 1+2\sum_{n\in\mathbb N}a_n. $$

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