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Let $a_n$ be a positive real sequence such that the series $\sum a_n$ converges.

I was asked to prove that under such circumstances $\sum a_n^{\frac{n}{n+1}}$ converges.

The previous sum can be rewritten as $\sum \frac{a_n}{a_n^{1/(n+1)}}$.

I can't prove the convergence of this last series... I'm sure comparison test is enough here, but other than $a_n \rightarrow 0$ there is no information that helps.

Thanks for your help.

Yiorgos S. Smyrlis
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Gabriel Romon
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4 Answers4

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Let $n$ be a positive integer.

If $a_n \le \frac{1}{2^{n+1}}$, then $a_n^{\frac{n}{n+1}}\le \frac{1}{2^n}$.

If $a_n \gt \frac{1}{2^{n+1}}$, then by your calculation $a_n^{\frac{n}{n+1}}\le 2a_n$.

Thus for all $n$ we have $a_n^{\frac{n}{n+1}}\le \frac{1}{2^n}+2a_n$.

Let $b_n=\frac{1}{2^n}+2a_n$. Then $\sum b_n$ converges, and therefore by Comparison so does our sum.

user84413
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André Nicolas
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  • Well that works, but how did you have the idea to check whether $a_n \leq 1/2^{n+1}$ or not ? – Gabriel Romon Feb 16 '14 at 17:23
  • Well, I looked at your expression, and noticed it is helpful if $a_n^{1/(n+1)}$ is not too small, but not helpful if $a_n^{1/(n+1)}$ is close to $0$. However, if that happens, then $a_n$ is small, so things are good then also. The rest was mechanical. – André Nicolas Feb 16 '14 at 17:27
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We have that by AM-GM

$$a_n^{\frac{n}{n+1}}\le \frac{2a_n+2a_n+\ldots+2a_n+\frac1{2^n}}{n+1}=\frac{2na_n+\frac1{2^n}}{n+1} \le2a_n+\frac1{2^n}$$

user
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    Nice application of the AM-GM (+1) – Mark Viola Aug 25 '20 at 15:58
  • @MarkViola Thanks! At first I tried with $na_n+1$ then with $a_n+2a_n+\ldots+na_n+\frac1{n!}$ and finally I had the ispiration from the answer by Mindlack to try with that. – user Aug 25 '20 at 16:00
  • Of course the number $2$ is not a magic number here. And I'm not sure the reason for choosing it. One could have said for any $C>1$, $a_n^{n/(n+1)}\le \frac{Cna_n+\frac1{C^n}}{n+1}$ – Mark Viola Aug 25 '20 at 16:04
  • @MarkViola Yes of course, any $C>1$ works fine! – user Aug 25 '20 at 16:09
  • @MarkViola What is the asymptotic for $a_n^{\frac{n}{n+1}}$? – user Aug 25 '20 at 16:10
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Hint: if $a_n \leq 2^{-n-1}$, then $a_n^{n/(n+1)} \leq 2^{-n}$. Otherwise, $a_n^{n/(n+1)} \leq 2a_n$.

Aphelli
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Let $A=\sum_{n=1}^\infty a_n$, and set $$S=\big\{n: a_n^{n/(n+1)}\le 2a_n\big\},$$ and $$T=\big\{n: a_n^{n/(n+1)}> 2a_n\big\}.$$

If $n\in T$, then $$ a^{n/(n+1)}_n> 2a_n\quad\Longrightarrow\quad a_n^n>2^{n+1}a_n^{n+1}\quad\Longrightarrow\quad a_n<2^{-n-1}. $$

$$ \sum_{n=1}^\infty a^{n/(n+1)}_n= \sum_{n\in S} a^{n/(n+1)}_n+\sum_{n\in T} a^{n/(n+1)}_n\le 2\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty 2^{-n-1}=2\sum_{n=1}^\infty a_n+\frac{1}{2}. $$ Thus $$ \sum_{n=1}^\infty a^{n/(n+1)}_n\le \frac{1}{2}+2A<\infty. $$

Yiorgos S. Smyrlis
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