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]
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.