Sum of independent r.v. converge in distribution to Poisson(1)

Question

Let $X_{n1}, X_{n2}, . . . , X_{nn}$ be independent random variables with a common distribution given as follows: $P(X_{nk} = 0) = 1 − 1/n-1/n^2$ , $P(X_{nk} = 1) = 1/n$, $P(X_{nk} = 2) = 1/n^2$ , where $k = 1, 2, . . . , n$ and n = $2, 3, . . . .$

Set $S_n = X_{n1} +X_{n2} +· · ·+X_{nn}$, $n ≥ 2$. Show that $S_n → \textrm{Poisson}(1)$ as $n → ∞$ in distribution.

I decided to check by characteristic functions convergence: Let $Y\sim \textrm{Poisson}(1)$

$E(e^{itY})=e^{(e^{it}-1)}.$

$E(e^{itS_n})=E(e^{it(X_{n1}+X_{n2}+...+X_{nn})})$, here are $n$ independent elements, so $E(e^{itS_n})=[E(e^{itX_{n1}})]^n$

$E(e^{itX_{n1}}) = e^{it0}0+e^{it}(1/n)+e^{2it}(1/n^2)=\frac{e^{it}}{n}(1+\frac{e^{it}}{n})$

as $n→∞$ $[\frac{e^{it}}{n}(1+\frac{e^{it}}{n})]^n→0$.

Which is wrong. Can you help me?

Answer 1

In your computation of $Ee^{itX_{n1}}$ you have taken $P\{X_{nk}=0\}$ as $0$ instead of $1-1/n-1/n^{2}$.