Proof using Stolz–Cesàro theorem I was asked to:
Applying Stolz–Cesàro theorem show the following result (named geometric mean). If $(u_n)$ converges to $L>0$ then $(u_1u_2u_3...u_n)^{1/n}, n \in \mathbb{N}$ also converges to $L$.
Despite of being able to derive the arithmetic mean result using Cesáro theorem, I've not be able to do the same for the geometric mean one.
 A: We assume that $u_k>0$ for all $k\in \mathbb{N}$.  Note that we can write
$$\left(\prod_{k=1}^n u_k\right)^{1/n}=e^{\frac1n \sum_{k=1}^n\log(u_k)}\tag 1$$
Next, the Stolz-Ceasro Theorem guarantees that 
$$\begin{align}
\lim_{n\to \infty}\frac1n \sum_{k=1}^n\log(u_k)&=\lim_{n\to \infty}\frac{\sum_{k=1}^{n+1}\log(u_k)-\sum_{k=1}^n\log(u_k)}{(n+1)-n}\\\\
&=\lim_{n\to \infty}\log(u_{n+1})\\\\
&=\log(L)\tag 2
\end{align}$$
Finally, using the result from $(2)$ in $(1)$ along with continuity of the exponential function yields the coveted limit
$$\lim_{n\to \infty}\left(\prod_{k=1}^n u_k\right)^{1/n}=L$$
A: From $AM-GM-HM$ we get 
$$\frac{u_1+u_2+\dots+u_n}{n} \geq (u_1u_2\cdots u_n)^{1/n}\geq \frac{n}{\frac{1}{u_2} +\frac{1}{u_2} +\cdots \frac{1}{u_n}}$$
You say you know $\frac{u_1+u_2+\dots+u_n}{n}\rightarrow L$,
Also,
\begin{align*}
\lim_{n\rightarrow \infty}\frac{n}{\frac{1}{u_2} +\frac{1}{u_2} +\cdots \frac{1}{u_n}}&=^{CS}\lim_{n\rightarrow \infty}\frac{n+1 -n}{\frac{1}{u_{n+1}}} \\
&=L
\end{align*}
And  by the squize theroem we conclude.
