We are given $X_n\sim B(n,p_n)$ where $np_n\rightarrow\lambda$, and $\lambda>0$.
The goal is to prove $X_n$ converges in distribution to Poisson($\lambda$) by use of characteristic functions.
My issue is proving this with $p_n$ being arbitrary as it is. Here are my current steps:
$\phi_{X_n}(t)=(1-p_n+p_ne^{it})^n=(1+p_ne^{it}-p_n)^n=(1+\frac{np_n(e^{it}-1)}{n})^n=e^{n\ln(1+\frac{np_n(e^{it}-1)}{n})}=e^\frac{\ln(1+\frac{np_n(e^{it}-1)}{n})}{1/n}$
Then evaluating the limit means examining the exponent:
$\displaystyle\lim_{n\rightarrow\infty}\frac{\ln(1+\frac{np_n(e^{it}-1)}{n})}{1/n}=\frac{\ln(1+\frac{\lambda(e^{it}-1)}{\displaystyle\lim_{n\rightarrow\infty} n})}{\displaystyle\lim_{n\rightarrow\infty}1/n}=\frac{\ln(1)}{0}=\frac{0}{0}$.
At this point I look to applying L'Hoptials rule, but I need to tweak something because when doing a u-substitution of $u = 1+\frac{np_n(e^{it}-1)}{n}$ I'm not sure how to evaluate the derivative to get the result.
I would imagine it is just zero, treating $p_n$ as a constant, but then this falls apart. As for treating the derivative with $p_n$, I don't see the trick. My first thought was the difference quotient manually, but that did not lead anywhere obvious.
The result should be $\displaystyle\lim_{n\rightarrow\infty}\phi_{X_n}(t)=e^{\lambda(e^{it}-1)}$, which is the $Poisson(\lambda)$ characteristic function.
Any feedback is much appreciated.
Best Answer
At the crucial step you have (writing $\lambda_n=np_n$) $$\ln \phi_n(t)= n \ln \left(1 + \frac {\lambda_n (e^{it}-1)}n\right)$$ which you should expand as $$\ln \phi_n(t)= n \left(\frac {\lambda_n (e^{it}-1)}n + o( \frac {\lambda_n (e^{it}-1)}n) \right)$$ which is $$\ln \phi_n(t)= \lambda_n (e^{it}-1)+ o(\lambda_n (e^{it}-1)) $$ which converges to $\lambda(e^{it}-1)$ as $n\to\infty$ and $\lambda_n\to\lambda$.
If you like you can write the expansion $\ln (1+x) = x + \theta x^2/2$, where $|\theta|\le 1$ when you plug in $x=\lambda_n (e^{it}-1)$ here, for added comfort.