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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?

Jose M Serra
  • 2,583
scobaco
  • 597

2 Answers2

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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 $n \to \infty$.

RRL
  • 90,707
1

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 $0$.

Lutz Lehmann
  • 126,666