$X_n \sim \text{Exponential}(\lambda_n)$, independent, $\sum 1/\lambda_n = \infty$, then, $\sum X_n=\infty$ a.s.

Let $\{X_n\}$ be a sequence of independent Exponential random variables with mean $$E(X_n)=\frac{1}{\lambda_n},$$ where $$0 < \lambda_n < \infty.$$ If $$\sum \frac{1}{\lambda_n} = \infty,$$ show $$P\left(\sum X_n = \infty\right) = 1.$$

It looks like a job for the Borel-Cantelli lemma, but I don't see which sequence of events will give the result.

Let $$X=\sum X_n,$$ then $$\sum \frac{1}{\lambda_n}=E(X) \ge \sum P(X>n),$$ but this is in the wrong direction.

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1. The event $\sum_j X_j=\infty$ is a tail event, hence its probability is $0$ or $1$.
2. Assume it is $0$; then the series $\sum_j X_j$ is almost surely convergent, hence it is supposed to converge in law.

3. Computing the characteristic function of partial sums, and in particular their modulus, one can see this can't happen.

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Whoa! Of very great level. Thanks. –  Nicolas Essis-Breton Apr 29 '13 at 20:25
You are welcome. –  Davide Giraudo Apr 29 '13 at 20:30