Characteristic function of the Binomial distribution converges to that of the Poisson > Find conditions on $\lambda, n, p$, so that the characteristic function of the Binomial converges to that of the Poisson > > Binomial distribution is given as $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ > > Poisson distribution is given as $P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$ So characteristic function of binomial: $\phi_X(t)=E(e^{itX})=\sum_1^n \binom{n}{k} e^{itk} p^k(1-p)^{n-k}=\sum \binom{n}{k} (e^{it}p)^k (1-p)^{n-k}=(pe^{it}+(1-p))^n$ and that of Poisson in a similar way:$\quad$$e^{\lambda(e^{it}-1)}$ Now my questions: 1) We defined Expectation with integral, why here with a sum ? 2) Can you give me a hint to solve the problem ?

Concerning your questions

1. Expectation is defined as an integral in the continuous case. Now you have discrete random variables. The discrete analogon of the integral is the sum (actually vice versa, i.e. the integral denotes an infite sum with infinitesimal (instead of integer valued) increment). So, for random variables with **discrete** values you will have sums, and for random variables that take values in **continuous** intervals you will have integrals.
2. Write the characteristic function of the binomial as follows $$(pe^{it}+(1-p))^n=\left(1+\frac{np(e^{it}-1)}{n}\right)^n$$ Denote $np$ with $λ$ and use the fact that $$\lim_{n \to \infty}\left(1+\frac xn\right)^n=e^x$$ to conclude as required.