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?

Your approach is in fact OK if you compute more carefully (although characteristic function method is always preferable).
Given $N = n$, $S$ has Gamma distribution $\Gamma(n, \mu)$ (see here), i.e., the conditional density of $S$ given $N = n$ is 
$$f_{S|N = n}(s) = \frac{\mu^n}{\Gamma(n)}s^{n - 1}e^{-\mu s}, \quad s > 0. $$
Therefore the marginal cdf of $S$ can be computed as
\begin{align*}
& P(S \leq x) = \sum_{n = 1}^\infty P(S \leq x|N = n)P(N = n) \\
= & \sum_{n = 1}^\infty\left(\int_0^x \frac{\mu^n}{\Gamma(n)}s^{n - 1}e^{-\mu s} ds\right) \times (1 - p)p^{n - 1} \\
= & \mu(1 - p) \int_0^x \left(\sum_{n = 1}^\infty\frac{(\mu p s)^{n - 1}}{(n - 1)!}\right)e^{-\mu s}ds \\
= & \mu(1 - p)\int_0^x e^{\mu p s}e^{-\mu s} ds \\
= & \mu(1 - p) \int_0^x e^{-\mu(1 - p)s} ds \\
= & 1 - e^{-\mu(1 - p)x},
\end{align*}
which shows that $S \sim \exp(\mu(1 - p))$. The interchange of $\int$ and $\sum$ follows from the Fubini's theorem.
