$X_n$ is an iid binary random process with equal probability of +1 or -1 occurring at any time $\displaystyle n.Y_n=\frac{1}{\sqrt{n}}\sum ^{n-1}_{k=0}X_k $,and i had already found that its characteristic function is $e^{n\ln\cos(\frac{u}{\sqrt{n}})}$.
Find the limit of this expression as $n \to +\infty$,why isn't the answer is $1$?here is my thinking:
n $\to \infty$ means $\cos(\frac{u}{\sqrt{n}}) \to 1$,then $\log 1=0$,so $e^{n\ln\cos(\frac{u}{\sqrt{n}})}$ become $e^{\infty*0}=e^{0}=1$,
However, the answer is $e^{\frac{-u^2}{2}} $,here is the formula(solution) of $e^{\frac{-u^2}{2}} $
$f(u)=\ln\cos(\frac{u}{\sqrt{n}})=f(0)+f'(0)u+\frac{f''(0)}{2}u^2+....$,$f'(0)=0;f''(0)=\frac{-1}{n}$,so $f(u)=0-\frac{1}{2n}u^2+\frac{1}{n^{\frac{3}{2}}}u^3$,so $n\ln\cos(\frac{u}{\sqrt{n}})=n(-\frac{u^2}{2n}+...)=\frac{-u^2}{2}+0(n^{-0.5}+...)$,so when n $\to \infty$,the characteristic function will become $e^{\frac{-u^2}{2}}$
I don't know where am i wrong,my answer is 1,but teacher's answer is $e^{\frac{-u^2}{2}}$,can anyone help me?