Convergence of averging series If $\sum\limits_{n=1}^\infty a_n$ converges where $a_n>0,\ \forall n\in\mathbb N$, prove $\sum\limits_{n=1}^\infty \sqrt[n]{a_1a_2\cdots a_n}$ and $\sum\limits_{n=1}^\infty\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$ converge.
$\sum\limits_{n=1}^\infty \frac{\sum\limits_{i=1}^na_i}{n}$ diverges, so one can not use the bound of the GM-AM relation at least directly.
 A: Since $$\displaystyle  \sqrt[n]{\prod_{i=1}^n a_i} \geq \frac{n}{\sum_{i=1}^n a_i^{-1}}$$ it suffices to prove the convergence of $\displaystyle \sum\limits_{n=1}^\infty \sqrt[n]{a_1a_2\cdots a_n}$.
This follows directly from Carleman's inequality
A: Consider $b_n=\frac{1}{a_n}$, $\sum\limits_{n=1}^\infty\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$ converges is equal to $\sum\limits_{n=1}^\infty\frac{n}{b_1+b_2+\cdots+b_n}$ converges.
$Firstly,$if $b_{n+1}\geqslant b_n$for any $b_i$, we have
$b_1+\cdots+b_{2n-1}\geqslant b_n+\cdots+_{2n-1}\geqslant nb_n$
$$\Longrightarrow\frac{2n-1}{b_1+\cdots+b_{2n-1}}+\frac{2n}{b_1+\cdots+b_{2n}}\leqslant\frac{2n-1}{nb_n}+\frac{2n}{nb_n}\leqslant\frac{4}{b_n}=4a_n$$
$$\Longrightarrow\sum_{n=1}^\infty\frac{n}{b_1+b_2+\cdots+b_n}\leqslant4\sum_{n=1}^{\infty}\frac{1}{b_n}$$
$Secondly,$ arrange {$b_n$}, make it increase, and then we get {$c_n$},for any $n$,we have $$c_1+c_2+\cdots+c_n\leqslant b_1+b_2+\cdots+b_n$$
$$\Longrightarrow\frac{n}{b_1+b_2+\cdots+b_n}\leqslant\frac{n}{c_1+c_2+\cdots+c_n}$$
$\quad\sum\limits_{n=1}^{\infty}\frac{n}{c_1+c_2+\cdots+c_n}$ converges , so $\sum\limits_{n=1}^{\infty}\frac{n}{b_1+b_2+\cdots+b_n}$ converges.
Add:
since $a_n\to 0$ , for any $\varepsilon>0$, exist a $N$,while $n>N$, we have $a_n<\varepsilon$, similarly, for any $A>0$, exist a $N$,while $n>N$, we have $b_n>A$, that is to say, there are finite number of $b_i\leqslant A$, so I can take out the  minimum and the second minimum and so on.
Besides, for I have proved that $\sum\limits_{n=1}^{\infty}\frac{n}{b_1+b_2+\cdots+b_n}$ converges, after my arrangement of it, it's limit will not change, so $\sum\limits_{n=1}^{\infty}\frac{n}{c_1+c_2+\cdots+c_n}=\sum\limits_{n=1}^{\infty}\frac{n}{b_1+b_2+\cdots+b_n}$.
According to $Firstly$ and $Secondly$, We get $\sum\limits_{n=1}^\infty\frac{n}{b_1+b_2+\cdots+b_n}$ converges.
$viz.\sum\limits_{n=1}^\infty\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$ converges.
Another Carleman's inequality,my proof is similar to LeGrandDODOM's hyperlink.
