Example of nonnegative random variables $X_n$ such that $\sum\limits_{n\ge1}X_n$ converges a.s. but $\sum\limits_{n\ge1}EX_n$ diverges. 
Utilize series of the form $\sum\limits_{n\ge1}\frac{1}{n^p}$ to construct independent, nonnegative random variables $X_n$ such that $\sum\limits_{n\ge1}X_n$ converges a.s. but $\sum\limits_{n\ge1}EX_n$ diverges.

I am quite stumped on this one. I know $X_n=n\cdot\mathbb{1}_{(0,\frac{1}{n})}$ is a typical example of random variables such that
\begin{align*}
\sum_{n\ge1}EX_n=\sum_{n\ge1}n\cdot P\big(\big(0,\frac{1}{n}\big)\big)=\sum_{n\ge1}(1)=\infty
\end{align*}
However these random variables are not independent and I am not sure that $\sum\limits_{n\ge1}X_n$ converges a.s. If we let $A_n$ be disjoint intervals of length $\frac{1}{n}$ and set $X_n=n\cdot\mathbb{1}_{A_n}$, then the $X_n$ are independent this time and $\sum_{n\ge1}EX_n=\infty$ again, as above. But I am not sure that $\sum\limits_{n\ge1}X_n$ converges a.s., if they do, is there an nice way to see this? Any help with this or any other example of $X_n$'s that will satisfy the required properties would be greatly appreciated.
 A: Let $X_n:= n^\alpha \mathbf{1}_{A_n}$, where $(A_n)_{n\geqslant 1}$ is a sequence of independent sets and $A_n$ has probability $p_n$, with $\alpha$ and $p_n$ specified later. If $\sum_{n\geqslant 1}p_n$ converges, so does $\sum_{n\geqslant 1}X_n$, by using Borel-Cantelli lemma. Note that $EX_n=n^\alpha p_n$ hence we can choose $p_n=n^{-2}$ and $\alpha =2$ for example.
For the construction of the sequence of sets, one can work on an infinite product of the unit interval endowed with the Lebesgue measure.
A: Consider on some probability space a sequence of independent random variables $(X_n)_{n\in\mathbb N}$ such that
\begin{align*}
X_n=
\begin{cases}
n&\text{with probability $\dfrac{1}{n^2}$,}\\
0&\text{with probability $1-\dfrac{1}{n^2}$;}
\end{cases}
\end{align*}
for a general construction of the underlying probability space, see Theorem 20.4 in Billingsley (1995, p. 265).
For each $m\in\mathbb N$, let
\begin{align*}
E_m\equiv\{X_n=0\text{ for every $n\geq m$}\}.
\end{align*}
By independence, one has
\begin{align*}
\mathbb P(E_m)=\prod_{n=m}^{\infty}\left(1-\frac{1}{n^2}\right)=\frac{m-1}{m}.
\end{align*}
Define $E\equiv\bigcup_{m=1}^{\infty} E_m$. Since $E_1\subseteq E_2\subseteq E_3\subseteq\cdots$, it follows that
\begin{align*}
\mathbb P(E)=\lim_{m\to\infty}\mathbb P(E_m)=1.
\end{align*}
But on $E$, only finitely many members of $(X_n)_{n\in\mathbb N}$ can be positive, so
\begin{align*}
\sum_{n=1}^{\infty}X_n<\infty.
\end{align*}
At the same time,
\begin{align*}
\sum_{n=1}^{\infty}\mathbb E(X_n)=\sum_{n=1}^{\infty}\left[(n)\left(\frac{1}{n^2}\right)+(0)\left(1-\frac{1}{n^2}\right)\right]=\sum_{n=1}^{\infty}\frac{1}{n}=\infty.
\end{align*}

My first answer was wrong, as pointed out by @RobertIsrael. I just realized this amended answer is essentially the same as the much simpler one by @DavideGiraudo (once you exploit the Borel–Cantelli lemma).
