$X_1$, $X_2$,..., are independent random variables. $X_k$ has pdf
$$f_k(x) =\frac{x^k}{k!}e^{−x}I_{(0,\infty)}(x)$$
Let
$$T_n=\frac{\sum_{i=1}^n(2X_i−n)}{n}−3$$
Consider the sequence $T_1$, $T_2$,.., and give the pmf or pdf of the limiting distribution.
I tried to use the following theorem:
$$\lim_{n\rightarrow\infty}M_{T_n}(t)=M_T(t)$$ for all $t$ in a neighborhood of $0$ $\Rightarrow T_n \rightarrow T$
I first note that $f_k(x)\sim Gamma(\alpha=k+1,\beta=1)$ which has mgf $M_{X_k}(t)=\left(\frac{1}{1-t}\right)^{k+1}$
Then
$$\begin{align*} M_{T_n}(t) &=M_{\frac{\sum(2X_i−n)}{n}−3}(t)\\\\ &=M_{\frac{2\left(\sum X_i\right)−n^2}{n}−3}(t)\\\\ &=M_{\frac{2}{n}\left(\sum X_i\right)-n−3}(t)\\\\ &=e^{-(n+3)t}M_{\sum X_i}\left(\frac{2t}{n}\right)\\\\ &=e^{-(n+3)t}M_{X_1}\left(\frac{2t}{n}\right)\cdot M_{X_2}\left(\frac{2t}{n}\right)\cdots M_{X_n}\left(\frac{2t}{n}\right)\\\\ &=e^{-(n+3)t}\cdot\left(\frac{1}{1-\frac{2t}{n}}\right)^{2}\cdot \left(\frac{1}{1-\frac{2t}{n}}\right)^{3}\cdots \left(\frac{1}{1-\frac{2t}{n}}\right)^{n+1}\\\\ &=e^{-(n+3)t}\cdot\left(\frac{1}{1-\frac{2t}{n}}\right)^{\frac{(n+1)\cdot(n+2)}{2}-1} \end{align*}$$
Then the next step would be to take
$$\lim_{n\rightarrow\infty}\left(e^{-(n+3)t}\cdot\left(\frac{1}{1-\frac{2t}{n}}\right)^{\frac{(n+1)\cdot(n+2)}{2}-1}\right)$$
which looks rather messy. I am afraid I made a mistake somewhere. Is there a more simple approach to finding the limiting distribution?
I have also tried taking
$$\begin{align*} F_{T_n}(t) &=P(T_n\leq t)\\\\ &=P\left(\frac{\sum(2X_i−n)}{n}−3\leq t\right)\\\\ &=P\left(\sum X_i \leq \frac{n(t+3)+n^2}{2}\right)\\\\ \end{align*}$$
but this method wouldn't work as the $X_k$'s are not identically distributed.
Method 1. Using what you have observed, realize each $X_k$ as the sum of $k+1$ i.i.d. exponential random variables $Y_{k,0}$, $\cdots$, $Y_{k,k}$ of the unit rate. If we denote $S_n = \sum_{k=1}^{n} X_k = \sum_{j,k : 0\leq j \leq k \leq n} Y_{k,j}$, then $S_n$ is the sum of $\frac{n(n+3)}{2}$ i.i.d exponential random variables and hence
$$ T_n = \frac{S_n - \frac{n(n+3)}{2}}{\frac{n}{2}} = \frac{S_n - \mathbf{E}[S_n]}{\sqrt{\mathsf{Var}(S_n)}} \cdot \sqrt{ \frac{2(n+3)}{n} }. $$
By the classical CLT, this converges in distribution to the normal distribution $\mathcal{N}(0, 2)$.
Method 2. Alternatively, starting from OP's computation and using the Taylor approximation
$$-\log(1-x) = x + \frac{x^2}{2} + \mathcal{O}(x^3),$$
we find that
\begin{align*} e^{-(n+3)t} \bigg( \frac{1}{1-\frac{2t}{n}} \bigg)^{\frac{(n+1)(n+2)}{2}-1} &= \exp\left\{ -(n+3)t + \frac{n(n+3)}{2}\left( \frac{2t}{n} + \frac{2t^2}{n^2} + \mathcal{O}(n^{-3}) \right) \right\} \\ &= \exp\left\{ \frac{n+3}{n} t^2 + \mathcal{O}(n^{-1}) \right\} \\ &\xrightarrow[n\to\infty]{} e^{t^2}, \end{align*}
which is the MGF of the normal distribution $\mathcal{N}(0,2)$.