Prove that $\lim_{n\to \infty} (a_1a_2\ldots a_n)^{\frac 1n} = L$ given that $\lim_{n\to \infty} (a_n) = L$ Let $\{a_n\}$ be a sequence of positive numbers such that $\lim_{n\to \infty} a_n = L$. Prove that $$\lim_{n\to \infty} (a_1a_2\ldots a_n)^{\frac 1n} = L$$
(Also given HINT: Let $a>0$. Then $\lim_{n\to\infty} a^{\frac 1n}=1$)
What I have so far:  $$\text{Let}\:\epsilon>0.\:\text{Then}\;\exists\;N\in\mathrm{I}\!\mathrm{N}\;\text{such that}\;n\ge N.$$$$\text{Then}\;L-\epsilon\lt a_n \lt L + \epsilon$$
$$\text{Let}\; b_n =(a_1a_2\ldots a_n)^{\frac 1n} \text{for}\; n\ge N$$
$$\text{Now}\;b_n = (a_1a_2\ldots a_N)^{\frac 1n}(a_{N+1}\ldots a_n)^{\frac 1n}$$
I am struggling with how to use the given hint and maybe limit superior/inferior facts to prove the limit is L. Any tips would be greatly appreciated. Also this is my first time using MathJax so please let me know if I have made any mistakes. Thank you.  
EDIT: I'm looking specifically on how to prove it using $\limsup$ or $\liminf$ properties!
 A: With the ideas of limsup and liminf in mind, we can conjure up an alternate approach than given in other answers.
Let $\{a_n\}$ be a positive sequence such that $a_n\rightarrow L$. Assume, to start, that $L>0$. Choose $\varepsilon$ such that $L>\varepsilon>0$. Then choose $N$ large enough so that for all $n>N$, we have that
$$L-\varepsilon\leq a_n\leq L+\varepsilon\,.$$
Finally, set $M=a_1\ldots a_N$ to save on space. Then we have that
$$\sqrt[n]{M}\sqrt[n]{(L-\varepsilon)^{n-N}}\leq\sqrt[n]{a_1\ldots a_n}=\sqrt[n]{Ma_{N+1}\ldots a_n}\leq \sqrt[n]{M}\sqrt[n]{(L+\varepsilon)^{n-N}}\,.$$
The inequalities on the left and right respectively give
$$L-\varepsilon\leq \liminf\sqrt[n]{a_1\ldots a_n}\quad\text{ and }\quad \limsup\sqrt[n]{a_1\ldots a_n}\leq L+\varepsilon\,.$$
Since $\varepsilon$ is arbitrary, we get our desired limit.
If $L=0$, we needn't worry about the inequality on the left. We can settle for the cruder estimate
$$0\leq \sqrt[n]{a_1\ldots a_n}$$
while still using liminf.
A: Note that
$$(a_1a_2\ldots a_n)^{\frac 1n}=e^{\frac{\sum\log a_i}{n}}$$
and by Stolz-Cesaro
$$\lim_{n\to \infty}\frac{\sum_i^n\log a_i}{n}=\lim_{n\to \infty} \frac{\sum_i^{n+1}\log a_i-\sum_i^{n}\log a_i}{n+1-n}=\lim_{n\to \infty} 
\log a_{n+1}=\log L$$
thus
$$\lim_{n\to \infty} (a_1a_2\ldots a_n)^{\frac 1n} = e^{\log L}=L$$
A: The trick is to convert the geometric mean to an arithmetic mean. See  here . Then find the limit of the arithmetic mean. See here.
A: First, for all $\epsilon>0$ there exists $n_0\in\mathbb{N}$ such that for all $n\geq n_0$ we have
$$
L-\epsilon< a_n< L+\epsilon\quad(1)
$$
We know that
$$
\frac{n}{\frac{1}{a_1}+\cdots+\frac{1}{a_n}}\leq(a_1a_2\cdots a_n)^{\frac{1}{n}}\leq\frac{a_1+\cdots+a_{n}}{n}\quad(2)
$$
Hence, we have
$$
\frac{n}{\frac{1}{a_1}+\cdots+\frac{1}{a_{n_0}}+\frac{(n-n_0)}{(L-\epsilon)}}\leq(a_1a_2\cdots a_n)^{\frac{1}{n}}\leq\frac{a_1+\cdots+a_{n_0}}{n}+\frac{(n-n_0)}{n}(L+\epsilon)\quad(3)
$$
Note that
$$
\lim_{n\to\infty}\left[\frac{a_1+\cdots+a_{n_0}}{n}+\frac{(n-n_0)}{n}(L+\epsilon)\right]=L+\epsilon
$$
and
$$
\lim_{n\to\infty}\frac{\frac{1}{a_1}+\cdots+\frac{1}{a_{n_0}}+\frac{(n-n_0)}{(L-\epsilon)}}{n}=\frac{1}{L-\epsilon}
$$
Then for all $\epsilon>0$
$$
L-\epsilon\leq\lim_{n\to\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}}\leq L+\epsilon
$$
