Central limit theorem for exponential random variable Consider the following problem:
Suppose that $(e_n)_n$ are independent exponentially distributed random variable with $E[e_n]=\mu_n.$ If $$\lim_n\max_{1 \leq k \leq n} \frac{\mu_k}{\sum_{r=1}^n\mu_r}=0.$$
then $$\sum_{k=1}^n(e_k-\mu_k)/\sqrt{\sum_{k=1}^n\mu^2_k} \implies N(0;1).$$
It's obvious that we need to prove Lindeberg condition, so in order to prove it, we need the assumption $\lim_n\max_{1 \leq k \leq n} \frac{\mu_k}{\sum_{r=1}^n\mu^2_r}=0$ and not the above.
Is it possible the there is a typo in this exercise? Is there any counterexample to disprove it?
Note that this question was asked before here, but it's not confirmed.
 A: Suppose $ \{X_n,n\ge 1\}$ are exponentially distributed random variables with $ \mathsf{E}[X_n]=\mu_n $. Then
\begin{equation*}
 \lim_{n\to\infty}\frac{\max\limits_{1\le i\le n}\mu_i}{\sum\limits_{j=1}^n \mu_j}=0 \tag{1}
\end{equation*}
is not sufficient for the following CLT
\begin{equation*}
 \frac{\sum\limits_{i=1}^n (X_i-\mu_i)}{\sqrt{\sum\limits_{i=1}^{n}\mu_j^2}}
 \Rightarrow N(0,1). \tag{2}
\end{equation*}
For example, $\mu_n=\frac1n$. In this case,
\begin{equation*}
 \max\limits_{1\le i\le n}\mu_i=1 , \qquad
 \lim_{n\to\infty}\sum_{j=1}^{n}\mu_j=\infty,
\end{equation*}
hence (1) holds.
On the other hand, since $ \mathsf{Var}[X_n]=\mu_n^2=\frac1{n^2} $,
\begin{equation*}
 c^2=\sum_{j=1}^{\infty} \mathsf{Var}[X_j]=\sum_{j=1}^{\infty}\frac1{j^2}\Big(=\frac{\pi^2}{6}\Big)<\infty.
\end{equation*}
Using Kolmogorov Convergence Criterion(Please refer to the book
"S. I. Resnick, A Probability Path, Birkhäuser Boston, 2005." Th.7.3.3, p212.
) to get both
\begin{equation*}
 S_1=\sum_{j=1}^{\infty}(X_j-\mu_j),\quad  S_2=\sum_{j=2}^{\infty}(X_j-\mu_j) 
\end{equation*}
converge almost surely.
Mean while $(X_1-\mu_1)$, $S_2  $ are independent mutually,  hence $ S_1=(X_1-\mu_1)+S_2 $
couldn't be Gaussian distributed and
\begin{equation*}
 \lim_{n\to\infty}\frac{\sum\limits_{i=1}^n (X_i-\mu_i)}{\sqrt{\sum\limits_{i=1}^{n}\mu_j^2}}
 =\frac{S_1}{c}\stackrel{d}{\ne}N(0,1).
\end{equation*}
This means that (1) is not sufficient for (2)
