Determine the distribution of the random variable $Y=\sum_{k=1}^{\infty}kX_k$ Fix $p \in (0,1)$ and consider independent Poisson random variables $X_k$, $k \geq 1$ with $\mathbb E[X_k]=\frac{p^k}{k}$. Verify that the sum $\sum_{k=1}^{\infty}kX_k$ converges with probability one and determine the distribution of the random variable $Y=\sum_{k=1}^{\infty}kX_k$.
So, by Kolmogrov's two-series theorem, I can show the almost sure convergence, but how to calculate the distribution of $Y$?
THank you!
 A: tl;dr: look at the characteristic function, invoke independence of the summands to get a product of characteristic functions, compute the closed-form expression, and finally squint hard at the resulting expression to recognize a known characteristic function.
(This is a good method whenever you have a r.v. defined as the sum of independent (possibly weighted) "nice" r.v.'s with simple distributions.)

Details: Consider the characteristic function: for $t\in\mathbb{R}$,
$$
\mathbb{E} e^{itX} 
=\mathbb{E} \exp\left(\sum_{k=1}^\infty itk X_k\right) 
=\mathbb{E} \prod_{k=1}^\infty e^{itk X_k} 
\operatorname*{=}_{(1)} \prod_{k=1}^\infty \mathbb{E}  e^{itk X_k} 
\operatorname*{=}_{(2)} \prod_{k=1}^\infty \exp\left(\frac{p^k}{k}\left(e^{itk}-1\right)\right)
$$
where (1) is by independence, and (2) from the expression of the characteristic function of a $\operatorname{Poisson}\left(\frac{p^k}{k}\right)$ r.v.
From there, for any $t\in\mathbb{R}$,
$$
\mathbb{E} e^{itX} 
= \prod_{k=1}^\infty \exp\left(\frac{p^k}{k}\left(e^{itk}-1\right)\right)
= \exp\left(\sum_{k=1}^\infty \frac{p^k}{k}\left(e^{itk}-1\right)\right) 
= \exp\left(\sum_{k=1}^\infty \frac{(pe^{it})^k}{k} - \sum_{k=1}^\infty \frac{p^k}{k}\right)
$$
and using the fact that, for $a\in\mathbb{C}$ satisfying $\lvert a\rvert < 1$ we have $$
\sum_{k=1}^\infty \frac{a^k}{k} = -\ln(1-a) \tag{$\dagger$}
$$
we obtain
$$
\mathbb{E} e^{itX} 
= \exp\left(\sum_{k=1}^\infty \frac{(pe^{it})^k}{k} - \sum_{k=1}^\infty \frac{p^k}{k}\right)
= \exp\left(- \ln(1-pe^{it}) + \ln(1-p)\right) 
= \frac{1-p}{1-pe^{it}}
$$
which turns out to be the expression of the characteristic function of a Geometric r.v. with parameter $1-p$ (or, equivalently, and a tad more contrived, of a Negative Binomial with parameters $1$ and $p$).
