Showing that a Gamma distribution converges to a Normal distribution Consider $G = \operatorname{Gamma}(p)$. As $p$ goes to $\infty$, the Gamma becomes more and more bell-shaped. How do I show that $\frac{G - p}{\sqrt{p}} \to Z \sim N(0,1)$ as $p \to \infty$? 
I started with the CDF of the Gamma and began taking the limit, but it got very messy.
 A: This answer is part of a previous answer with a link here. That portion of the previous answer is copied over here so that one can see that the question above has been answered, however, as the answer here formed only part of an answer to a different question, it might not have been noticed in the different context of the question above. Text as follows:
A more direct relationship between the gamma distribution (GD) and the normal distribution (ND) with mean zero follows. Simply put, the GD becomes normal in shape as its shape parameter is allowed to increase. Proving that that is the case is more difficult. For the GD, $$\text{GD}(z;a,b)=\begin{array}{cc}
& 
\begin{cases}
 \dfrac{b^{-a} z^{a-1} e^{-\dfrac{z}{b}}}{\Gamma (a)} & z>0 \\
 0 & \text{other} \\
\end{cases}
\,. \\
\end{array}$$
As the GD shape parameter $a\rightarrow \infty$, the GD shape becomes more symmetric and normal, however, as the mean increases with increasing $a$, we have to left shift the GD by $(a-1) \sqrt{\dfrac{1}{a}} k$ to hold it stationary, and finally, if we wish to maintain the same standard deviation for our shifted GD, we have to decrease the scale parameter ($b$) proportional to $\sqrt{\dfrac{1}{a}}$.
To wit, to transform a GD to a limiting case ND we set the standard deviation to be a constant ($k$) by letting $b=\sqrt{\dfrac{1}{a}} k$ and shift the GD to the left to have a mode of zero by substituting $z=(a-1) \sqrt{\dfrac{1}{a}} k+x\ .$ Then $$\text{GD}\left((a-1) \sqrt{\frac{1}{a}} k+x;\ a,\ \sqrt{\frac{1}{a}} k\right)=\begin{array}{cc}
 & 
\begin{cases}
 \dfrac{\left(\dfrac{k}{\sqrt{a}}\right)^{-a} e^{-\dfrac{\sqrt{a} x}{k}-a+1} \left(\dfrac{(a-1) k}{\sqrt{a}}+x\right)^{a-1}}{\Gamma (a)} &  x>\dfrac{k(1-a)}{\sqrt{a}} \\
 0 & \text{other} \\
\end{cases}
 \\
\end{array}\,.$$
Note that in the limit as $a\rightarrow\infty$ the most negative value of $x$ for which this GD is nonzero $\rightarrow -\infty$. That is, the semi-infinite GD support becomes infinite. Taking the limit as $a\rightarrow \infty$ of the reparameterized GD, we find
$$\lim_{a\to \infty } \, \frac{\left(\frac{k}{\sqrt{a}}\right)^{-a} e^{-\frac{\sqrt{a} x}{k}-a+1} \left(\frac{(a-1) k}{\sqrt{a}}+x\right)^{a-1}}{\Gamma (a)}=\dfrac{e^{-\dfrac{x^2}{2 k^2}}}{\sqrt{2 \pi } k}=\text{ND}\left(x;0,k^2\right)$$
Graphically for $k=2$ and $a=1,2,4,8,16,32,64$ the GD is in blue and the limiting $\text{ND}\left(x;0,\ 2^2\right)$ is in orange, below

