Distribution of sum of random variables Let $X_1, X_2, . . .$ be independent exponential random variables with mean $1/\mu$ and let $N$ be
a discrete random variable with
$P(N = k) = (1 − p)p^{k-1}$  for $k = 1, 2, . . . $
where $0 ≤ p < 1$ (i.e. $N$ is a shifted geometric random variable). Show that $S$ defined as $S =\sum_{n=1}^{N}X_n$ is again exponentially distributed with parameter $(1 − p)\mu$.
My approach:
$S = \sum_{n=1}^{N}\sum_{k=1}^{\infty}\frac{1}{\mu}e^{-\frac{n}{\mu}}(1-p)p^{k-1}$
How to solve this? Is it the right approach to solve this problem?

I would calculate the characteristic function :
$$E[ e^{itS} ] = E\left[ \prod_{i=1^N} e^{itX_1} \right]$$
$$= \sum_{k=1}^\infty (1-p)p^k \prod_{i=1}^k \int_{\mathbb{R}^+} e^{itx} \mu e^{-\mu x} dx$$
$$= \sum_{k=1}^\infty (1-p)p^{k-1} \prod_{i=1}^k \frac{\mu}{\mu-it}$$
$$  =\frac{(1-p)}{p}\sum_{k=1}^\infty \left( \frac{ \mu p }{\mu - it} \right)^k$$
$$ = \frac{(1-p)}{p} \frac{ 1 }{ 1-  \frac{ \mu p }{\mu - it} } - (1-p)$$
$$= \frac{(1-p)}{p} \frac{\mu - it - (\mu - it -   \mu p)}{ \mu - it -   \mu p}$$
$$ = \frac{(1-p)\mu }{ (1-p) \mu - it}$$
And this is exactly the characteristic function pof an exponential variable with parameter $(1-p)\mu$
