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Given a positive sequence $(a_n)$. Prove that if $\displaystyle \sum_{n=1}^{\infty} a_n$ converges, then so does $\displaystyle \sum_{n=1}^{\infty} a_n^{\frac{n}{n+1}}$.

I tried to use the root test, and I was stuck with the case $\lim \sup \sqrt[n]{a_n}=1$. Maybe we need some trick here.

Thanks so much.

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  • $\begingroup$ A place to start might be $a_n^{n/(n+1)}=a_n/a_n^{1/(n+1)}$. $\endgroup$ – marty cohen Feb 15 '15 at 6:22
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Let $b_n=a_n+e^{-n}$. The specific choice $e^{-n}$ here isn't too important, we just want to make sure that the $b_n$'s still converge but not too quickly. Now $\sum b_n$ converges and $b_n^{n/(n+1)}>a_n^{n/(n+1)}$, so if $\sum b_n^{n/(n+1)}$ converges then so does $\sum a_n^{n/(n+1)}$.

Now we can use the limit comparison test, in its one-directional form:

If $(a_n),(b_n)$ are positive sequences and $\liminf_{n\to\infty}\frac{a_n}{b_n}>0$, then if $(a_n)$ converges so does $(b_n)$.

$$\liminf_{n\to\infty}\frac{b_n}{b_n^{\frac n{n+1}}}=\liminf_{n\to\infty}b_n^{\frac 1{n+1}},$$ so taking a $\log$, we wish to show that $\liminf_{n\to\infty}\frac{\log(b_n)}{n+1}>-\infty.$ But

$$b_n>e^{-n}\implies \liminf_{n\to\infty}\frac{\log(b_n)}{n+1}>\liminf_{n\to\infty}\frac{-n}{n+1}=-1.$$

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