TITLE: $\sum E(\xi_i) = \infty$ imply $\sum\xi_i =\infty$ a.s. for nonnegative independent $\xi_n$
QUESTION [8 upvotes]: Let $\xi_1,\xi_2\cdots$ be nonnegative, independent random variables with $\xi_n$ having distribution $\lambda_n \exp(-\lambda_n x)dx$, $x\geq0$ and $\lambda_n>0$. Assume $\sum_{n=1}^\infty \lambda_n^{-1}=\sum_{n=1}^\infty E(\xi_n)=\infty$, show that $\sum_{n=1}^\infty \xi_n = \infty$ a.s.

This is a homework problem, I'd found the same problem here( 
$X_n \sim \text{Exponential}(\lambda_n)$, independent, $\sum 1/\lambda_n = \infty$, then, $\sum X_n=\infty$ a.s.), but I don't have characteristic function as tool yet, I don't know how otherwise to check $P\{\sum_{n=1}^\infty\xi_n = \infty\} \neq 0$ after using zero- one law.
BTW, I will accept characteristic function method though if no other clean proof is found(I don't know how step 3 in the above link gives contradiction after obtaining the characteristic function.)

REPLY [3 votes]: A cheap trick is to use domination of a symmetric variable. Define
$$X_n=\begin{cases}
0&\text{ if $\xi_n\leq\lambda_n^{-1}\ln(2)$}\\
\lambda_n^{-1}\ln(2)&\text{ otherwise.}
\end{cases}$$
Since $\lambda_n^{-1}\ln(2)$ is the median of $\xi_n,$ each variable $X_n$ is symmetrically distributed about its mean: it has probability $\tfrac 1 2$ of being zero and probability $\tfrac 1 2$ of being $\lambda_n^{-1}\ln(2).$
Since $\sum_{n=1}^\infty\lambda_n^{-1}$ diverges, there is a sequence $0=n_1<n_2<\dots$ with $\sum_{n=n_k+1}^{n_{k+1}}\lambda_n^{-1}\ln(2)\geq 2$ for each $k.$
Since a sum of independent symmetric variables is symmetric, the random variables $Y_k=\sum_{n=n_k+1}^{n_{k+1}}X_i$ are symmetric, and the mean of $Y_k$ is at least $1$ so $\mathbb P[Y_k\geq 1]\geq \tfrac 1 2.$ The event that $Y_k\geq 1$ occurs for infinitely many $k$ has probability $1$ by the Borel-Cantelli lemma. And in that event, $\sum_{n=1}^\infty \xi_n\geq \sum_{k=1}^\infty Y_k$ must be infinite.