Limit of geometric mean of series terms. Suppose an infinite series converges,  $\sum a_n = S > 0,$  and $a_n > 0.$ 
I know that (arithmetic-geometric inequality) $0 < (\prod_\limits{i=1}^n a_i)^{1/n} < (\sum_\limits{i=1}^n a_i) / n.$
This shows that $\lim_\limits{n \to \infty}(\prod_\limits{i=1}^n a_i)^{1/n} = 0$ since $\lim_\limits{n \to \infty}(\sum_\limits{i=1}^n a_i) / n = S/\infty = 0.$
I want to show 
$$\lim_{n \to \infty}n(\prod_{i=1}^n a_i)^{1/n} = 0.$$
I can't continue because $0 < n(\prod_\limits{i=1}^n a_i)^{1/n} < (\sum_\limits{i=1}^n a_i)$  has a limit of the upper bound equal to $S$
Suggestions?
 A: You can still use the AM-GM inequality, but a more subtle approach is needed. 
Consider,
$$nG_n = \left(\frac{n^n}{n!}\right)^{1/n}\left( n!\prod_{k=1}^na_k\right)^{1/n} =  \left(\frac{n^n}{n!}\right)^{1/n}\left( \prod_{k=1}^nka_k\right)^{1/n}.$$
The AM-GM inequality now shows 
$$0 \leqslant nG_n \leqslant \left(\frac{n^n}{n!}\right)^{1/n}\frac{1}{n} \sum_{k=1}^nka_k.$$
We have $(n^n/n!)^{1/n}  \to e$ as $n \to \infty$. This follows from Cauchy's second limit theorem:
$$\lim_{n \to \infty} \left(\frac{n^n}{n!}\right)^{1/n} = \lim_{n \to \infty} \frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n} = \lim_{n \to \infty} \left(1 + 1/n\right)^n = e$$
By Kronecker's lemma,
$$\tag{1}\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^nka_k = 0,$$
and by the squeeze theorem it follows that $nG_n \to e \cdot 0 =0$ as $n \to \infty$.
A: Split the product in the middle and apply arithmetic-geometric mean:
$$
\left(\prod_{k=1}^{2n}a_k\right)^{\frac1{2n}}
=\left(\prod_{k=1}^{n}a_k\right)^{\frac1{2n}}
  ·\left(\prod_{k=n+1}^{2n}a_k\right)^{\frac1{2n}}
\le\sqrt{\frac1n\sum_{k=1}^{n}a_k}·\sqrt{\frac1n\sum_{k=n+1}^{2n}a_k}
$$
Now
$$
(2n)\left(\prod_{k=1}^{2n}a_k\right)\le 2·\sqrt{S}·\sqrt{\sum_{k=n+1}^{2n}a_k}
$$
and by the Cauchy property of converging series, the last factor converges to $0$.
