If $a_n>0$ converges to $a>0$, then $(a_0 a_1\cdots a_n)^{\frac{1}{n}}$ converges to $a.$ I would appreciate your help!
How can we show that if a sequence of positive real numbers $a_n$ converges to $a\in\mathbb{R}$ with $a>0$, then $(a_0 a_1\cdots a_n)^{\frac{1}{n}}$ converges to $a$ as well.
 A: Look at $\exp (\frac{1}{n} \sum_{m=0}^n \log a_m)$ and since $\log a_m\rightarrow \log a$ (assuming $a_n,a\neq 0$), you can apply the Cesaro mean. 
A: Using the following facts


*

*If a sequence of positive terms converges to a positive limit, then its Arithmetic Mean also converges to the same limit (for a proof see here).

*If A.M. (Arithmetic Mean) of a positive sequence converges, then the G.M. (Geometric Mean) of that sequence also converges to the same limit (this can be obtained by observing the fact that  A.M.  $\ge$ G.M.$\ge$ H.M.).

*If $M>0$ then the sequence $M^\frac{1}{n}\to 1$
one can obtain the required result. 
For details see below. 

Step-1: So if $a_n>0$ and $a_n\to a$ for some $a>0$, then $a_n$ is a bounded sequence i.e. $\exists\, M>0$ such that $a_n\le M$ for all $n\ge 0$. Let $A_n=\frac{1}{n+1}\sum\limits_{k=0}^n{a_k}$, $G_n=\left(\prod\limits_{k=0}^n{a_k}\right)^{\frac{1}{n+1}}$ and $H_n=\dfrac{n+1}{\frac{1}{a_0}+\dotsm+\frac{1}{a_n}}$. Then clearly $A_n\to a$. Also using the same argument, applying to the sequence $\frac{1}{a_n}$ one can obtain $H_n\to a$. Since $H_n\le G_n\le A_n$ for all $n\ge 0$, we have $|H_n-a|\le |G_n-a|\le |A_n-a|$, for all $n\ge 0$. Hence by the squeeze theorem, we have $G_n\to a$.
Step-2: Now let $b_n=\left(\prod\limits_{k=0}^n{a_k}\right)^\frac{1}{n}$. We now prove that $b_n\to a$. Choose $\varepsilon>0$. As $G_n$ is convergent, there is a $M^\prime>0$ such that $G_n\le M^\prime$ for all $n\ge 0$ and also there is a $n_1\in\Bbb N$ such that $|G_n-a|<\frac{\varepsilon}{2}$ for all $n\ge n_1$. Since $M^\frac{1}{n}\to 1$,  there is a $n_2\in \Bbb N$ such that $\left|M^\frac{1}{n}-1\right|<\frac{\varepsilon}{2M^\prime}$ for all $n\ge n_2$. Choose $n_0=\max\{n_1,n_2\}$. So if $n\ge n_0$, we have 
\begin{eqnarray}
|b_n-a| & \le & |G_n-a|+|b_n-G_n|\\
&<& \frac{\varepsilon}{2}+ G_n\left|\left(\prod\limits_{k=0}^n{a_k}\right)^\frac{1}{n(n+1)}-1\right|\\
&\le& \frac{\varepsilon}{2}+M^\prime\left|M^\frac{1}{n}-1\right|\quad\quad(\text{since}\,\, a_n\le M)\\
&<& \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.
\end{eqnarray}

A: WLOG $a \neq 0$. Suppose $N \in \mathbb{N}$ such that for $n \geq N$, $|a_n| > \epsilon$. 
Then for  $n \geq N$ know that:
$\log(a_n) \to \log(a) \implies \sum_{k=N}^{\infty} \log(a_n)=\log(a) \iff (a_Na_{N+1}...a_{N+k})^{\frac{1}{k}} \to a$.
You should be able to figure out the rest. (Note you will need that $a_n \neq 0 \forall n$. In fact, assuming this, you can show that you can take $N=0$ in the above.).
