I was wondering the following problem:
If $X_n$ is a sequence of independent random variables, $|X_n|\leq b, (b>0)$ and $X_n$ converges to $X$ almost surely. Using dominated convergence theorem, we have $\mathbb{E}[\lim\limits_{n\rightarrow \infty}X_n] = \lim\limits_{n\to\infty}\mathbb{E}[X_n] = \mathbb{E}[X]$. Similarly, we have $\mathbb{E}[\lim\limits_{n\rightarrow \infty}X_n^2] = \lim\limits_{n\to\infty}\mathbb{E}[X_n^2] = \mathbb{E}[X^2]$. Therefore, we have $Var(\lim\limits_{n\rightarrow \infty}X_n) = \lim\limits_{n\to\infty}Var(X_n) = Var(X)$
Then, using cesaro sum we have $\lim\limits_{n\to\infty}\frac{\sum^n_{i=1} X_i}{n} = X$ almost surely. So, $Var(\lim\limits_{n\to\infty}\frac{\sum^n_{i=1} X_i}{n}) = Var(X) = \lim\limits_{n\to\infty}Var(\frac{\sum^n_{i=1} X_i}{n}) = \lim\limits_{n\to\infty}\frac{1}{n^2}Var(\sum^n_{i=1} X_i) = \lim\limits_{n\to\infty}\frac{\sum^n_{i=1}Var(X_i)}{n^2} $.
On the other hand, using cesaro sum for the sequence $Var(X_n)$, we have $\lim\limits_{n\to\infty}\frac{\sum^n_{i=1}Var(X_i)}{n} = Var(X)$. Combine with the previous results, we have $\lim\limits_{n\to\infty}\frac{\sum^n_{i=1}Var(X_i)}{n} = Var(X)= \lim\limits_{n\to\infty}\frac{\sum^n_{i=1}Var(X_i)}{n^2}$.
This is obviously not correct, but I don't know where went wrong. Any help would be greatly appreciated!