The Problem
We have sequence of independent random variables $(X_n)_{n\ge1}$ ~ $U[\frac1n,1]$. I want to prove that $$\frac{X_1+X_2+...+X_n}{n}$$ converges almost surely and find the limit.
Theorem's I used
1. If $(X_n)_{n\ge1}$ is sequence of independent random variables , $S_n=X_1+X_2+...+X_n$,
and $\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2} <+\infty$, then $\frac{S_n-E(S_n)}{n} \rightarrow 0$ almost surely.
2. $X$ ~ $U[a,b] \Rightarrow Var(X)=\frac{(b-a)^2}{12}$
3. $X$ ~ $U[a,b] \Rightarrow E(X)=\frac{a+b}{2}$
My Justification
We know that $Var(X_n)=\frac{(1-\frac1n)^2}{12}=\frac{(n-1)^2}{12n^2}$
$\sum_{n=1}^{\infty}\frac{Var(X_n)}{n^2}=\sum_{n=1}^{\infty}\frac{(n-1)^2}{12n^4}<\sum_{n=1}^{\infty}\frac{n^2}{12n^4}<+\infty$.
$E(S_n)=E(X_1+X_2+...+X_n)=E(X_1)+E(X_2)+...+E(X_n)=1+\frac{3}{4}+\frac{4}{6}+...+\frac{n+1}{2n}$.
So by my theorem I see, that :
$$\frac{X_1+X_2+...+X_n}{n} \rightarrow \frac{1+\frac{3}{4}+\frac{4}{6}+\frac{5}{8}+...+\frac{n+1}{2n}}{n}$$ almost surely.
Is that correct ? I am little bit confused because of having on the right side number's dependent on $n$.