# If $X_n$ are i.i.d. $Uniform(0,1)$ then show that $S_n$ converges a.s. to $\infty$

Suppose $\{X_n\}$ is an i.i.d. $\text{Uniform}(0,1)$ sequence of random variables. Define $S_n=X_1+X_2+...+X_n$. Show that $S_n\to\infty$ almost surely.

I believe I have solved the problem and I wish the community would check if what I have done is sufficient, and general, i.e. for any i.i.d. sequence of random variables, my method would work.

We first consider, for $\varepsilon\in(0,1)$, $P(X_n>\varepsilon)=1-\varepsilon$ for all $n\in\mathbb N$ hence $\lim_{n\to\infty}P(X_n>\varepsilon)=1-\varepsilon\neq0$. So $X_n$ does not converge in probability to $0$. So $X_n$ does not converge almost surely to $0$. So $S_n$ diverges almost surely.

EDIT: $S_n$ is an increasing sequence of non-negative terms and we have shown that $S_n$ diverges almost surely. Hence $S_n\to\infty$ almost surely.

• This proves that $(S_n)$ diverges, and does it neatly, but not that $S_n\to\infty$ almost surely. – Did May 25 '15 at 16:06
• Is there a difference between the two? $S_n$ is a sequence of nonnegative random variables. It will diverge if and only if it goes to infinity. What am I missing? – Landon Carter May 25 '15 at 16:29
• Then add this to the proof. – Did May 25 '15 at 16:33
• LOL. You had me there!! :D – Landon Carter May 25 '15 at 16:38