Let $X_n \sim \text{Gamma}(n, \lambda)$, with $n \in \mathbb N$ and $\lambda > 0$. Let $Y_n = a_n(X_n - b_n)$ with $a_n, b_n \in \mathbb R$. Find $a_n$ and $b_n$ such that $Y_n$ converges in distribution to a standard normal.
Convergence in distribution is equivalent to convergence of characteristic functions; we have to prove that $$\Phi_{Y_n}(t) \to \Phi_{\mathcal N(0, 1)}(t) = e^{-\frac12 t^2}$$
We have $$\Phi_{X_n}(t) = {\left(\frac{\lambda}{\lambda - it}\right)}^n$$ and since $Y_n = a_n X_n - a_n b_n$ $$\Phi_{Y_n}(t) = e^{-i a_n b_n t} \Phi_{X_n}(a_n t) = e^{-i a_n b_n t} {\left(\frac{\lambda}{\lambda - i a_n t}\right)}^n$$
We will find $a_n, b_n$ such that $\log\Phi_{Y_n}(t) \to \log\Phi_{\mathcal N(0, 1)}(t) = -\frac12 t^2$. Let's assume for a moment that $|a_n| < |\lambda / t|$: this allows us to expand the logarithm in $(1)$. We will verify this assumption later. We have: $$\require{cancel}\begin{align*} \log\Phi_{Y_n}(t) &= -i a_n b_n t + n\left(\cancel{\log \lambda} - \cancel{\log\lambda} - \log(1 - i a_n t)\right) = \tag1\\ &= -i a_n b_n t - n\left(-i\frac{t}{\lambda}a_n - i^2\frac{t^2}{\lambda^2} a_n^2 + \mathcal O(a_n^3)\right) =\\ &= -i a_n b_n t + in\frac{t}{\lambda}a_n -\frac{t^2}{\lambda^2} n a_n^2 + \mathcal O(na_n^3) \end{align*}$$
Now, the first two terms has to cancel out or go to zero, the third one must converge to $-\frac12t^2$ and it has to be $\mathcal O(n a_n^3) \to 0$.
From the third one, we deduce $$a_n = \frac{\lambda}{\sqrt{2n}}$$ and this also satisfies the fourth condition. Condition $(1)$ also holds eventually. Since the second term does not go to zero, the only hope is that the first two terms cancel out. Plugging in $a_n$ we find $$b_n = \frac{n}{\lambda}$$ and we are done.
Question: I'm fairly confident the proof is correct, but I'd like to know for sure. On the other hand, I'm wary of its length. The question comes from a past exam and maybe I'm missing something simple to solve it in a shorter way?
EDIT: A now deleted answer suggested to use the CLT. I think that's the correct way to solve the problem, but I get a different answer for $a_n$ and $b_n$. Since the Gamma distribution, with a natural number as first parameter, is a sum of exponentials, we have that $X_n = \sum_{i = 1}^n W_i$, with $W_i \sim \text{Exp}(\lambda)$. Now, applying the CLT, $$\sqrt n(\bar W_n - \lambda^{-1}) \xrightarrow{d} \mathcal N(0, \lambda^{-2})$$ therefore $$\sqrt{n}\lambda(\bar W_n - \lambda^{-1}) \xrightarrow{d} \mathcal N(0, 1)$$ Since $\bar W_n = \frac1n X_n$, with a bit of algebra we find $$\frac{\lambda}{\sqrt n}\left(X_n - \frac{n}{\lambda}\right) \xrightarrow{d} \mathcal N(0, 1)$$ giving $a_n = \lambda / \sqrt{n}$ and $b_n = n/\lambda$. But $a_n$ is different than what I obtained above. So where's the error?