$X_n$ are independent exponential variables, if $\sum_{n=1}^\infty λ_n^{-1}=\infty$,show that $\sum_{n=1}^\infty X_n=\infty$ a.e. Let $X_1,X_2,\cdots,X_n$ be independent nonnegative random variables, with $X_n$ having density $λ_n\exp(-λ_n x),x\geq 0,λ_n\geq 0$,if $\sum_{n=1}^\infty λ_n^{-1}=\infty$, show that $\sum_{n=1}^\infty X_n=\infty$ a.e.
Attempts:$\sum_{n=1}^\infty λ_n^{-1}=\infty$ implies $\sum_n E(X_n)=E(\sum_n X_n)=\infty$,also I get a hint to try to consider $\exp(-\sum_nX_n)$,and from $X_n$ are independent,$E(\exp(-\sum_nX_n))=\Pi_n E(\exp(-X_n))$, but I don't know what to do next?
A very similar question:Let $X_1, X_2, \ldots$ be independent r.v.'s with $0 \leq X_n \leq 1$ and $\sum_n E(X_n) = \infty$. Show $\sum_n X_n = \infty$ with probability 1? but differs in $0\leq X_n\leq 1$
 A: The sum $\sum_{n=1}^\infty X_n$ converges if and only if $\sum_{n=1}^\infty \lambda_n^{-1}$ does. If $\sum_{x=1}^\infty X_n<\infty$ then for all $s>0$ we have
$$
0<\mathbb E\left[e^{-s\sum_{n=1}^\infty X_n}\right] = \prod_{n=1}^\infty \mathbb E[e^{-sX_n}] = \prod_{n=1}^\infty \frac{\lambda_n}{\lambda_n+s}.
$$
This infinite product converges if and only if $\sum_{n=1}^\infty \left(1-\frac{\lambda_n}{\lambda_n+s}\right)$ does (as can be shown by considering the convergence of $\sum_{n=1}^\infty -\log\left(\frac{\lambda_n}{\lambda_n+s}\right)$ ), and
$$
\sum_{n=1}^\infty \left(1-\frac{\lambda_n}{\lambda_n+s}\right) = \sum_{n=1}^\infty \frac s{s+\lambda_n}
$$
converges if and only if $\sum_{n=1}^\infty \lambda_n^{-1}$ does.
Conversely, if $\sum_{n=1}^\infty \lambda_n^{-1}<\infty$, then by monotone convergence, $$\mathbb \sum_{n=1}^\infty \mathbb E[X_n] = \mathbb E\left[\sum_{n=1}^\infty X_n\right]<\infty,$$ and hence $\sum_{n=1}^\infty X_n<\infty$ as the $X_n$ are almost surely nonnegative.
