$X_n$ $\,{\buildrel a.s. \over \rightarrow}\,$ $X$, then $(\prod_{i=1}^{n}X_i)^{1/n}$ $\,{\buildrel a.s. \over \rightarrow}\,$ $X$? Prove or provide a counterexample:
If $X_n$ $\,{\buildrel a.s. \over \rightarrow}\,$ $X$, then $(\prod_{i=1}^{n}X_i)^{1/n}$ $\,{\buildrel a.s. \over \rightarrow}\,$ $X$
My attempt:
FALSE:  Suppose $X$ can take on only negative values, and suppose $X_n \equiv X$   $\forall$ $n$ 
THEN $X_n$ $\,{\buildrel a.s. \over \rightarrow}\,$ $X$, however for even $n$, $(\prod_{i=1}^{n}X_i)^{1/n}$ is not strictly negative.  Instead, it alternates negative to posotive and negative.  Therefore, $(\prod_{i=1}^{n}X_i)^{1/n}$ does not converge almost surely to $X$.
Is this a reasonable answer??  If not, how can I improve my answer?
 A: Before proving something of interest, notice that $X_i >0$ almost surely for all $i$ is not a necessary condition for both statements to make sense, which the deterministic sequence $(-1, -1, 1, 1, 1, \dots)$ illustrates.
Moreover, the statement is indeed false in general, as the following deterministic sequence proves: $(0, 1, 1, \dots)$.
Now, suppose $X_i >0$ almost surely for all $i$, then the statement is true by the following argument:
Define $$S_n = \frac{1}{n}\sum_{i=1}^n\log(X_i).$$ By contuity of  $x\mapsto \log(x)$, $\log(X_n)\to\log(X)$ almost surely. Thus, $S_n \to\log(X)$ almost surely by a result for Cesaro means also proven in the comments above. Thus, by continuity of $x\mapsto \exp(x)$, $$\left(\prod_{i=1}^nX_i\right)^{1/n}\to X,$$ almost surely.
A: This claim is false. I give proof by providing a counterexample.
Suppose the random sequence $X_i$ is defined as follows:
\begin{align}
Z_i &\sim N(0,1/i), iid, \; \forall i \in \mathbb{N} \\
X_i &= 1_{\{i \neq 1\}} + 1_{\{i \neq 1\}}Z_i, \; i \in \mathbb{N}
\end{align}
Clearly, $X_i$ is (1) degenerate and (2) converges almost surely to $X=1$ as $i \longrightarrow \infty$ by Chebyshev's strong law of large numbers. (To see this, rewrite $Z_i = i^{-0.5}Z$ for $Z \sim N(0,1)$.) 
However, since $X_1 = 0$, $\Pi_{i=1}^nX_i = 0, \; \forall n \in \mathbb{N}$. Consequently, $(\Pi_{i=1}^nX_i)^{1/n} = 0, \forall n \in \mathbb{N}$, so it will in the limit trivially converge to $0$, that is $lim_{n\longrightarrow \infty}(\Pi_{i=1}^nX_i)^{1/n} = 0$. $\square$
