Fix $p\in(0,1)$ and consider independent Poisson random variables. $X_k$, $k\geq1$ with $E[X_k]=\frac{p^k}{k}$. Verify that the series $\sum\limits_{k=1}^{\infty}kX_k$ converges with probability one and determine the distribution of the random variable $Y=\sum\limits_{k=1}^{\infty}kX_k.$
I found $E[s^{X_k}]=\exp\left(\frac{p^k}{k}(s-1)\right)$, $E[s^{kX_k}]=\exp\left(\frac{p^k}{k}(s^k-1)\right)$ and defining $Y_n=\sum\limits_{k=1}^nkX_k$, I got $$E[s^{Y_n}]=\exp\left(\sum_{k=1}^n\frac{(sp)^k}{k}-\sum_{k=1}^n\frac{p^k}{k}\right).$$ Then taking limit as $n\rightarrow\infty$, I obtained $$E[s^Y]=\frac{1-p}{1-sp}.$$ By using the probability generating function of Y, I obtained $P(Y=k)=(1-p)p^k$ for $k=0,1,2,...$. This means that $Y_n=\sum\limits_{k=1}^{\infty}kX_k$ converges to random variable defined above in distribution. How can I verify that this sum converges with probability one?
We have $E[kX_k]=p^k$ and $\sum_{k=1}^{\infty}p^k=\frac{p}{1-p}<\infty$. Then, $Var(kX_k)=kp^k$ and $\sum_{k=1}^{\infty}kp^k=\frac{p}{(1-p)^2}<\infty$. So, by Kolmogorov's Two Series Test, the series $\sum_{k=1}^{\infty}kX_k$ converges almost surely.