Let $\{a_n\}$ be a sequence of positive numbers such that $\lim_{n\to\infty} a_n = L$. Prove that $$\lim_{n\to\infty}(a_1\cdots a_n)^{1/n} = L$$
Proof:
Let $\epsilon > 0$. There exists $N\in\mathbb{N}$ such that if $n\ge N$ then $L-\epsilon < a_n < L + \epsilon$. Now, let $b_n = (a_1\cdots a_n)^{1/n} $. We can split this up based on the tail of $\{a_n\}$: $b_n = (a_1\cdots a_{N})^{1/n} (a_{N+1}\cdots a_{n})^{1/n}$. We can bound $(a_{N+1}\cdots a_{n})^{1/n}$ by other sequence which have limits, since we have that $L-\epsilon < a_n < L + \epsilon$ for all $n\ge N$: $$(L-\epsilon)^{1-N/n} < (a_{N+1}\cdots a_{n})^{1/n} < (L+\epsilon)^{1-N/n}.$$ Let $C=a_1\cdots a_N$. If we multiply this inequality through by $C^{1/n}$ we find that $$C^{1/n}(L-\epsilon)^{1-N/n} < b_n < C^{1/n}(L+\epsilon)^{1-N/n}.$$ If we take the limit of this inequality, we find that $L - \epsilon \le \lim_{n\to\infty} b_n \le L+\epsilon$. So $L \le \lim_{n\to\infty}b_n \le L$. Hence $\lim_{n\to\infty}b_n = L$.
Does this argument work?