Sum of independent Poisson random variables Let $\{X_n: n \in \mathbb N\} $ be independent Poisson variables with respective parameters; $P(X_n = k) = e^{-\lambda_n} \lambda_n^k/k!$ ; $k = 0, 1, \ldots$.
Why is it valid that $ \sum(X_n) $ converges or diverges almost surely according as $\sum(\lambda_n) $ converges or diverges? 
 A: Since the Poisson distribution is supported on the set $\{0,1,2,3,\ldots\}$, a sum of them converges if and only if only finitely many of them are positive.  The sum of their probabilities of being positive is
$$
\sum_{n=0}^\infty \left( 1 - e^{-\lambda_n} \right).\tag{1}
$$
Notice that $\lambda \ge 1-e^{-\lambda}$ and for $\lambda$ near $0$, this is $\ge\lambda/2$, and $\lambda_n$ will be near $0$ if $n$ is large enough.  Hence by comparison, $(1)$ converges if and only if $\displaystyle\sum_{n=0}^\infty\lambda_n$ converges.  Now apply the two Borel–Cantelli lemmas.
A: *

*In this context, $S_n$ has Poisson distribution of parameter $\sum_{j=0}^n\lambda_j$.

*It's not hard to determine convergence in law of a sequence of random variables taking their values on a countable set.
Note that in this case, almost sure convergence is not hard to determine because the involves random variables take integer values. A good exercise will be to establish a similar result for example when $X_i\sim N(0,\sigma_i^2)$ (normal distribution). You can use the following:

If $\{Y_n\}$ are independent random variables, the series $\sum_{n=0}^{+\infty} Y_n$ is almost surely convergent if and only if it is convergent in law, that is, the sequence of partial sums $(S_n,n\in\Bbb N):=\left(\sum_{j=0}^nX_j,n\in\Bbb N\right)$ is almost surely convergent (respectively in law).

A: The sum of two independent Poisson random variables with rates $\lambda_1$ and $\lambda_2$ is Poisson distributed with rate $\lambda_1 + \lambda_2$. (See http://en.wikipedia.org/wiki/Poisson_distribution or you can calculate it yourself). From this  you can see that if $\sum \lambda_i$ is infinite, there is no hope of convergence, since the resulting random variable will have mean "infinity"...i.e. it diverges.
On the other hand if $\sum \lambda_i$ is finite, you can use a coupling argument to see that the convergence is almost sure. (See http://en.wikipedia.org/wiki/Coupling_(probability) for this technique if you havent seen it before) One way to create the Poisson random variable is to count the number of points in a Possion process with rate $\lambda$ on some interval, say $[0,1]$ for convenience, where the number of points per length rate is $\lambda$. If we put independent copies of Poisson random variables, then to compute the sum we count the TOTAL number of points. The rate per unit length will be a Possion random variable with rate $\sum \lambda_i$ which we can see since the number of points per unit length is $\sum \lambda_i$, the sum of the rates from each of the constituant $\lambda_i$.
