$\sum E(\xi_i) = \infty$ imply $\sum\xi_i =\infty$ a.s. for nonnegative independent $\xi_n$ 
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.)
 A: 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.
A: Applying Kolmogorov's three-series theorem immediately gives a proof.
Putting jokes aside, here is a rather crude solution. I am sure there will be a shorter and neater solution, but my brain has almost stopped working...

Write $\beta_n = 1/\lambda_n$ for simplicity and write $T_n = \lambda_n \xi_n$. It is easy to check that $(T_n)$ are i.i.d. and have the common distribution $\operatorname{Exp}(1)$. Also we write $S_n = \sum_{k=1}^{n} \beta_k$.


*

*Assume first that $\beta_n / S_n \not\to 0$. Then there exist $\epsilon > 0$ and $(n_k)$ such that $\beta_{n_k} \geq \epsilon S_{n_k}$ for all $k$. Then
$$ \sum_{j=1}^{n_k} \xi_j = \sum_{j=1}^{n_k} \beta_j T_j \geq \epsilon S_{n_k} T_{n_k} $$
and since $\mathbb{P}(T_{n_k} \geq 1 \text{ i.o.}) = 1$ by the second Borel-Cantelli's lemma, we have $\sum_{j=1}^{n_k} \xi_j \uparrow \infty$ with probability one. This proves the desired claim.

*Now assume that $\beta_n / S_n \to 0$. On the one hand, for $\epsilon \in (0, 1)$ Chebyshev's inequality yields
$$ \mathbb{P}\left( \sum_{j=1}^{n} \xi_j \leq \epsilon S_n \right)
= \mathbb{P}\left( \sum_{j=1}^{n} \beta_j(1 - T_j) \geq (1-\epsilon) S_n \right)
\leq \frac{1}{(1-\epsilon)^2 S_n^2} \sum_{j=1}^{n} \beta_j^2. $$
On the other hand, by the Stolz-Cesaro theorem applied to $\frac{\beta_n^2}{S_n^2 - S_{n-1}^2} = \frac{\beta_n}{S_n + S_{n-1}} \to 0$, it follows that $\frac{1}{S_n^2} \sum_{j=1}^{n} \beta_j^2 \to 0$. Then by extracting a subsequence $(n_k)$ so that
$$ \sum_{k=1}^{\infty} \frac{1}{S_{n_k}^2} \sum_{j=1}^{n_k} \beta_j^2 < \infty $$
is satisfied, Borel-Cantelli's lemma tells that
$$ \mathbb{P}\left( \sum_{j=1}^{n_k} \xi_j >  \epsilon S_{n_k} \text{ eventually} \right) = 1 - \mathbb{P}\left( \sum_{j=1}^{n_k} \xi_j \leq  \epsilon S_{n_k} \text{ i.o.} \right) = 1. $$
Therefore the claim follows also in this case.
