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I have to prove that if $\sum_{n=1}^{\infty} a_{n}$ is a convergent series with positive real numbers, then $\sum_{n=1}^{\infty} (a_{n})^\frac{n}{n+1}$ converges. I also wonder if the converse is true. Any suggestion, hint will
be very welcome. Thanks.

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    $\begingroup$ for the converse note that at the end of the series you have the following inequality $ a_n \leq a_n^{\frac{n}{n+1}}$ $\endgroup$ – clark Jun 15 '12 at 16:10
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if $a_n \geq \frac{1}{2^{n+1}} \Rightarrow a_n^{\frac{n}{n+1}} =\frac{a_n}{a_n^{\frac{1}{n+1}}} \leq 2a_n$ if $a_n \leq \frac{1}{2^{n+1}}$ then $a_n^{\frac{n}{n+1} } \leq \frac{1}{2^n}$. Therefore $a_n^{\frac{n}{n+1} } \leq 2a_n + \frac{1}{2^n}$ and by comparison test you are done.

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