# If $\lim_{n\to\infty}a_n=a$ and $a_n>0$ for all $n$, then we have $\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}=a$? [duplicate]

The question is motivated by my answer to the question: Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$. In that answer, a lemma was used but I don't remember where I saw it. I'd like to recover a proof.

Question: How should I prove the following lemma?

Lemma. If $\lim_{n\to\infty}a_n=a$ and $a_n>0$ for all $n$, then we have $$\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}=a. \tag{*}$$

Attempt. I found that it suffices to show that $$\lim_{n\to\infty}\frac{a_1+\cdots+a_n}{n}=a$$ which turns out to be not easy for me. I tried directly manipulating by definition but failed. How should I go on?

• Ah, so your problem is to show $a_n \to a \Rightarrow \frac{1}{N}\sum_1^N a_n \to a$? – Daniel Fischer Aug 6 '13 at 0:05

Choose $N$ large enough so that $n \geq N$ gives $|a_n - a| < \epsilon$. Then \begin{align*}\bigg|\frac{a_1 + \cdots + a_n}{n} - a\ \bigg| &= \bigg|\frac{(a_1 - a) + \cdots + (a_n-a)}{n}\bigg| \\ &\leq \frac{|a_1 - a| + \cdots + |a_N - a|}{n} + \frac{(n-N)\epsilon}{n}\end{align*}. I might be off by one on the right term, but it doesn't matter, since you now take the limit as $n \rightarrow \infty$. The right term goes to $\epsilon$ and the left term vanishes.