Suppose $\lambda_{1}$, $\lambda_{2},\ldots\lambda_{n}\ldots$ be a monotone sequence of positive real numbers that go to infinity. Let $X_{n}$ be sequence of random variables with $\Gamma(\lambda_{n},1)$ distribution. The probability density function of $X_{n}$ is given by
$f_{n}(x)=$
\begin{cases}
\frac{x^{\lambda_{n}-1}\times e^{-x}}{\Gamma(\lambda(n))},\text{if $x\ge 0$;}\newline
0, &\text{otherwise}
\end{cases}
Denote $Y_{n}=\frac{(X_{n}-\lambda_{n})}{\sqrt{\lambda_{n}}}$
Prove that $Y_{n}$ converges in distribution to standard normal random variable as n goes to infinity.
I first found the characteristic function of $Y_{n}$ and I could also do the same for moment generating function and I tried to show that characteristic function of $Y_{n}$ converges to the characteristic function of the standard normal random variable which is $e^{(-1/2)t^{2}}$. To do that I used $\Phi_{aX+b}(t)=e^{ibt}\times\Phi_{X}(at)$ and I found $\Phi_{X_{n}}(t)=(1-it)^{-\lambda_{n}}$.From here I found the characteristic function as $\Phi_{aX_{n}+b}(t)=(1-\frac{it}{\sqrt{\lambda_{n}}})^{-\lambda_{n}}\times e^{it(-\sqrt{\lambda_{n}})}$.Now I am calculating that this characteristic function converges to $1$ which is not the characteristic function of standard normal distribution.
You want to calculate, for fixed real $t$, $$\lim_{\lambda\to\infty} \left(1-\frac{it}{\sqrt\lambda}\right)^{-\lambda}\exp(-\sqrt\lambda i t).$$ Take logarithms, so you now want to know $$\lim_{\lambda\to\infty} -\lambda\log\left(1-\frac{it}{\sqrt\lambda}\right)-\sqrt\lambda it.$$ Use the approximation $\log(1-x)=-x-x^2/2 + o(x^3)$, valid for small $x$, with the choice $x=it/\sqrt\lambda$. It's now just a matter of keeping track of powers of $\lambda$ in the approximation: $$\begin{align*} -\lambda\log\left(1-\frac{it}{\sqrt\lambda}\right)-\sqrt\lambda it&= -\lambda \left(-\frac{it}{\sqrt\lambda} -\frac{(it)^2}{2\lambda} + o(\lambda^{-3/2})\right) - \sqrt\lambda it\\&=-t^2/2+o(1/\sqrt\lambda)\end{align*}$$