Convergence of Exponential R.V.s? This is a question on a study guide to prepare for a final. I wanted to make sure that my solution is correct. 
Let $\{X_n\}$ be a sequence of independent random variables defined on the space $(\Omega, S,P)$ where $X_n\sim exp(\lambda_n)$. Show if $\sum_{n=1}^{\infty}\frac{1}{\lambda_n}$ diverges then $\sum_{n=1}^{\infty}X_n $ diverges a.e. 
Attempt: 
Will attempt to show that $Y_n=e^{-\sum_{i=1}^{n}X_i}$ converges a.e. to $0$ and the implication follows. 
Consider for a given $\epsilon > 0$, $P(\cap_{n=1}^{\infty}\cup_{k\ge n}|Y_k|> \epsilon)=\lim_{n\to\infty}P(\cup_{k\ge n}|Y_k|> \epsilon)$ by continuity from below. Then by countable subadditivity:
$\lim_{n\to\infty}P(\cup_{k\ge n}|Y_k|> \epsilon)\le\lim_{n\to\infty}\sum_{k\ge n}P(|Y_k|>\epsilon)$
Note, since $\{X_n\}$ are independent then $\sum_{i=1}^{n}X_i\sim exp(\sum_{i=1}^{n}\lambda_i)$
Then $P(Y_k>\epsilon)=P(e^{-\sum_{i=1}^{n}X_i}>\epsilon)=P(\sum_{i=1}^{n}X_i<log(\frac{1}{\epsilon}))=1-e^{-\sum_{i=1}^{n}\lambda_i log(\frac{1}{\epsilon})}$
Combining everything: 
$P(\cap_{n=1}^{\infty}\cup_{k\ge n}|Y_k|> \epsilon)\le\lim_{n\to\infty}\sum_{k\ge n}P(|Y_k|>\epsilon)=\lim_{n\to\infty}\sum_{k\ge n}1-e^{-\sum_{i=1}^{n}\lambda_i log(\frac{1}{\epsilon})}=1-\lim_{n\to\infty}\sum_{k\ge n}e^{-\sum_{i=1}^{n}\lambda_i log(\frac{1}{\epsilon})}=1-\lim_{n\to\infty}\sum_{k\ge n}\epsilon^{\sum_{i=1}^{n}\lambda_i}=1-\epsilon^{\sum_{i=1}^{\infty}\lambda_i}$ If I can show that $\sum_{i=1}^{\infty}\lambda_i=0.$ Then the conclusion follows. I am having trouble making that leap. Can anyone offer some advice or another method to prove this statment. 
 A: We show the contrapositive, that is, if $\sum_{n=1}^\infty X_n $ converges a.e. then $\sum_{n=1}^\infty \frac1{\lambda_n}<\infty$. For all $s>0$ we have
$$
0 < \mathbb E[e^{-s\sum_{n=1}^\infty X_n}] = \prod_{n=1}^\infty \mathbb E[e^{-sX_n}] = \prod_{n=1}^\infty \frac{\lambda_n}{\lambda_n+s}.
$$
Now, for a sequence of numbers $0<a_n<1$,
$$
\prod_{n=0}^\infty a_n > 0 \iff \sum_{n=0}^\infty -\log(1-b_n)<\infty > 0,
$$
where $b_n=1-a_n$. Recall the Taylor approximation
$$
-\log(1-x) = x + O(x^2),
$$
from which we see that 
$$
\sum_{n=0}^\infty -\log(1-b_n)<\infty \iff \sum_{n=0}^\infty b_n<\infty.
$$
Letting $a_n = \frac{\lambda_n}{\lambda_n+s}$, we find that
$$
\sum_{n=1}^\infty \left(1- \frac{\lambda_n}{\lambda_n+s}\right) = \sum_{n=1}^\infty \frac s{s+\lambda_n} <\infty.
$$
Hence $\lim_{n\to\infty}\frac s{s+\lambda_n}=0$ so that $\lim_{n\to\infty}\lambda_n=+\infty$. Now,
$$
\lim_{n\to\infty} \frac s{s+\lambda_n}\cdot\frac{\lambda_n}s = \lim_{n\to\infty} \frac{\lambda_n}{s+\lambda_n} = 1,
$$
and therefore
$$
\sum_{n=0}^\infty \frac s{s+\lambda_n} <\infty \iff \sum_{n=1}^\infty \frac1{\lambda_n}<\infty.
$$
