I'm trying to find an asymptotic distribution of the follwing random variable $$Z_n=\sum_{i=1}^n Y_i$$ where $Y_i = I[T_i<t]$ with $T_i \text{~} Gamma(i, \lambda)$.
Here $t$ is a fixed number.
My initial trial was using the Lyapunov CLT ( or Linderberg CLT). First I noticed that by using integral by parts, $$E[Y_i]=P[Y_i=1]=P[T_i<t]=\sum_{k=i}^\infty \frac{(t/\lambda)^ie^{-t/\lambda}}{i!}$$
which is the tail probabilty of the poisson random variable with mean $t/\lambda$.
Let $p_i:=E[Y_i], R_i:=Y_i-p_i, S_n:=\sum_{i=1}^nR_i, \sigma_n^2:=Var(S_n)=\sum_{i=1}^np_i(1-p_i)$.
I'm trying to show that $R_i$ satisfies the Lyapunov condition where
$$ \sigma_n^{-(2+\delta)}\sum_{i=1}^nE[|R_i|^{(2+\delta)}]\to0\ \text{as $n\to \infty$ }$$.
With $\delta=1$, I get $$E[|R_i|^3]=(1-p_i)^3p_i + p_i^3(1-p_i)^3\le p_i(1-p_i)$$ therefore $$\sigma_n^{-3}\sum_{i=1}^nE[|R_i|^{3}] \le \sigma_n^{-1}.$$
As long as I can show that $$\sigma_n \to \infty \ \text{as $n\to \infty$ }$$ I'm done with the proof. But I'm stuck with showing the divergence of $\sigma_n$.
In other words, I need to show that
$$ \sum_{i=1}^n p_i(1-p_i) \to \infty \ \text{as $n\to \infty$ }.$$
I sincerely ask to help me with the proof of the last statement. Or Am i doing something wrong?
Thank you.