Let $(X_i)_{i\geq1}$ be $i.i.d$ Bernoulli ($\lambda$) and $S_n = \sum_{i=1}^nX_i $. Find the limiting distribution of $S_n$ as $n \rightarrow \infty$.
My approach: Use characteristic functions. $$\phi_{S_n}(t) = E[e^{itS_n}] = E[e^{it \sum X_i}] = \prod_i E[e^{itX_1}] = (E[e^{itX_1}])^n = ((e^{it}-1)\lambda +1)^n$$
$$\underset{n\rightarrow \infty}{lim} \phi_{S_n}(t) = \underset{n\rightarrow \infty}{lim} ((e^{it}-1)\lambda +1)^n$$
How to proceed further? Is it correct to expect the characteristic function to converge to that of a normal distribution for some $\mu$ and $\sigma$?
Any hints are appreciated. Thanks.
it doesn't converges in distribution, because if it does, the convergence will be, also, a.s, which is false, since $\sum_nP(|X_n|>1/2)=+\infty,$ using the three series theorem, so in order to have a convergence in distribution, you can consider $\sqrt{n}(\dfrac{1}{n}\sum_{k=1}^nX_k-p)$ whoch converges in distribution to $N(0,p(1-p))$ (central limit theorem)