Consider a sequence $(X_k)_{k\in\mathbb{N}}$ of independent Bernoulli$(1/k)$ random variables.
Then, as $n\rightarrow\infty$, $$\mathbb{E}\left[\sum_{k=1}^nX_k\right]=\sum_{k=1}^n1/k\sim\log(n)$$ and $$ \text{Var}\left[\sum_{k=1}^nX_k\right]=\sum_{k=1}^n(1/k-1/k^2)\sim\log(n).$$ Question: Is it true that $$ \frac{\sum_{k=1}^nX_k-\log(n)}{\sqrt{\log(n)}} $$ converges in distribution to a standard normal random variable?
Attempted answer:
The generating function of $$ Y_n:=\frac{\sum_{k=1}^nX_k-\log(n)}{\sqrt{\log(n)}} $$ is $$ \mathbb{E}\left[z^{Y_n}\right]=z^{-\sqrt{\log(n)}}\prod_{k=1}^n\mathbb{E}\left[\left(z^{1/\sqrt{\log(n)}}\right)^{X_k}\right]. $$ But $$ \mathbb{E}\left[\left(z^{1/\sqrt{\log(n)}}\right)^{X_k}\right]=\left(z^{1/\sqrt{\log(n)}}+1-k\right)/k... $$ I'm not sure how to progress. Perhaps there is some taylor expansion argument which I am not seeing. Any help much appreciated.