how to prove that ${S_N\over E[S_N]}$ converges to an exponential distribution Suppose that $\{X_1,X_2,\ldots\}$ is a sequence of iid $L^1$-random variables such that $E[X_1]\neq 0$. Define for every $n$, 
$$
S_n=X_1+\cdots+X_n.
$$
Let $N$ be a geometric random variable such that 
$$
P(N=k) = q^{k-1}p,\quad k=1,2,\ldots,
$$
where $q=1-p$ and $p\in(0,1)$. Assume that $N$ and $\{X_1,X_2,\ldots\}$ are all independent. Show that as $p\to0$, 
$$
{S_N\over E[S_N]}
$$
converges in distribution to an exponential distribution with some rate $\lambda$, and identify $\lambda$.
 A: You are already on the right track and I complete the latter part for you.
Let $\mu = E[X_1]$ be the common mean and 
$ \varphi(t) = E[e^{itX_1}]$ be the common characteristic function
As stated above,
$$ E[S_N] = E[E[S_N|N]] = E[N\mu] = \frac {\mu} {p} $$
Now consider the characteristic function of $\displaystyle Z = \frac {S_N} {E[S_N]} = \frac {pS_N} {\mu}$:
$$ \begin{align*} \varphi_Z(t) & = E\left[e^{it\frac {pS_N} {\mu}}\right] \\
& = E\left[E\left[e^{it\frac {pS_N} {\mu}}|N\right]\right] \\
& = E\left[E\left[e^{i\frac {tp} {\mu} X_1}\right]^N\right] \\
& = E\left[\varphi\left(\frac {tp} {\mu} \right)^N\right] \\ 
& = \sum_{k=1}^{+\infty} \varphi\left(\frac {tp} {\mu} \right)^k (1 - p)^{k-1}p \\
& = p\varphi\left(\frac {tp} {\mu} \right) \sum_{k=1}^{+\infty} 
\left[\varphi\left(\frac {tp} {\mu} \right) (1 - p)\right]^{k-1} \\
& = \frac {\displaystyle p\varphi\left(\frac {tp} {\mu}\right)} 
{\displaystyle 1 - \varphi\left(\frac {tp} {\mu} \right) (1 - p) }
\end{align*}$$
Note the infinite geometric series converges as the characteristic function is bounded by 1. Since $X_1 \in \mathcal{L}^1$, $\varphi$ is differentiable and thus we can evaluate the limit via L'Hopital Rule:
$$ \begin{align*}
\lim_{p\to 0} \varphi_Z(t) &= \lim_{p\to 0} 
\frac {\displaystyle p\varphi\left(\frac {tp} {\mu}\right)} 
{\displaystyle 1 - \varphi\left(\frac {tp} {\mu} \right) (1 - p) } \\
& = \lim_{p\to 0} \frac {\displaystyle \varphi\left(\frac {tp} {\mu}\right) + p\varphi'\left(\frac {tp} {\mu}\right)\frac {t} {\mu}} 
{\displaystyle  \varphi\left(\frac {tp} {\mu} \right) - (1 - p)\varphi'\left(\frac {tp} {\mu}\right)\frac {t} {\mu} } \\
& = \frac {\varphi(0) + 0} {\displaystyle \varphi(0) - \varphi'(0)\frac {t} {\mu}} \\
& = \frac {1} {\displaystyle 1 - i\mu \frac {t} {\mu}} \\
& = \frac {1} {1 - it}
\end{align*}$$
which is the characteristic function of $\text{Exp}(\lambda = 1)$
