Let $N$ be a geometrically distributed random variable with parameter $\frac{1}{2}$, i.e., $\mathbb{P}(N=n)=(\frac{1}{2})^{n+1},n\geq0$, and let $Y_{1},Y_{2},...$ be independent identically distributed copies of an exponentially distributed random variable with mean 2.
What is the Laplace-Stieltjes Transform $\mathbb{E}[e^{-sZ}]$ of the random sum $Z=2Y_{1}+\sum_{i=1}^{N}Y_{i+1}$?
So I know that I can use the property of independence to get \begin{equation} \mathbb{E}[e^{-sZ}]=\mathbb{E}[e^{-2sY_{1}}]\mathbb{E}[e^{-s\sum_{i=1}^{N}Y_{i+1}}] \end{equation}
I know that $\mathbb{E}[e^{-2sY_{1}}]=\int_{0}^{\infty}e^{-2st}\frac{1}{2}e^{-\frac{1}{2}t}dt=\frac{1}{1+4s}$.
Now for $\mathbb{E}[e^{-s\sum_{i=1}^{N}Y_{i+1}}]$ I believe I need to assume that $N$ and $Y_{i+1}$ are independent. Then \begin{equation} \mathbb{E}[e^{-s\sum_{i=1}^{N}Y_{i+1}}]=\sum_{n=1}^{\infty}\mathbb{E}[e^{-s\sum_{i=1}^{n}Y_{i+1}}](\frac{1}{2})^{n+1}=\sum_{n=1}^{\infty}(\frac{1}{1+2s})^{n}(\frac{1}{2})^{n+1} \end{equation}
I am wondering whether my computation here is actually correct. Even if it is correct I am unsure how to simplify this series. Could someone help me by telling me where I went wrong or how I should proceed?
Your solution seems correct. The only issue I have is that in your description $N$ can take the value $0$, but as you sum from $1$ to $N$ this can lead to errors. It makes more sense to use a different description of the geometric distribution, in which $N$ stands for the number of trials up to and including a success, which then has distribution $\mathbb{P}(N = n) = (\frac{1}{2})^n$ for any $n \geq 1$.
Using that description the Laplace-Stieltjes transform is given by \begin{align*} \mathbb{E}[e^{-s\sum_{i = 1}^N Y_{i + 1}}] = \sum_{n = 1}^\infty (\frac{1}{1 + 2s})^n (\frac{1}{2})^n = \sum_{n = 1}^\infty (\frac{1}{2 + 4s})^n. \end{align*} This summation is a geometric series (https://en.wikipedia.org/wiki/Geometric_series) which only converges if $|\frac{1}{2 + 4s}| < 1$ and it converges to \begin{align*} \frac{1}{1 - \frac{1}{2 + 4s}} - 1, \end{align*} (the 1 appears because the geometric series starts at $n = 0$)