Estimate Gamma distribution using central limit theorem The random variable X has a Gamma distribution with parameters $ \alpha$ = na, and $ \beta$ = b , $a, b > 0$, and n a natural number different from zero.
What is the probability that X is strictly greater than $(n+1)ab$ for n big enough ? 
i start with: $$ X_i > (n+1)ab$$
so: $$X_i -nab > ab $$
now divide by standard deviation of Gamma distribution:$$\frac{X_i -nab}{\sqrt {nab^2} } > \frac{a}{\sqrt {na}} $$
and now take the sum over all $X_i$'s: $$\sum_{i=0}^n \frac{X_i -nab}{\sqrt {nab^2} } > \frac{na}{\sqrt {na}} $$
and for last i multiply $ \frac{1}{\sqrt n}$ with the term above: $$\frac{1}{\sqrt n}\sum_{i=0}^n \frac{X_i -nab}{\sqrt {nab^2} } > \frac{na}{\sqrt {a}n} $$
or equivalently:$$\sum_{i=0}^n \frac{X_i -nab}{\sqrt {a} nb } > \frac{a}{\sqrt {a}} $$
so i get exactly the structure of central limit theorem by manipulating the fact that $X_i > (n+1)ab$ that says it is equal to standard normal distribution the left hand side is now standard normal distributed , but clearly i make a mistake some where but i don't see it! Can  someone please explain why my method doesn't work ?
 A: I don't quite understand your question, but I'll give it a shot.
Let $X\sim\Gamma(na,b)$, $n\in\mathbb{N}_{>1};a,b\in\mathbb{R}_{>0}$. Now define iid $X_i\sim\Gamma(a,b)$, so that $$ X=\sum_{i=1}^n X_i $$
First, let $ \mu={a}{b} $ and $\sigma^2=ab^2$ (assuming shape-scale, not shape-rate, parametrization). 
Then, by the central limit theorem, 
 $$ Z = \frac{X - n\mu}{\sigma\sqrt{n}} $$
converges in distribution to the standard normal, i.e. $$ Z=\frac{X - nab}{b\sqrt{an}} \xrightarrow[\;\;\;]{\text{d}} \mathcal{N}\left(0,1\right) $$
However, it appears you are interested in the fact that the distribution of $X/n$ is close to $ \mathcal{N}(\mu,\sigma^2/n) $ for large enough $n$ (verbatim from Wiki). Please correct me if I'm misunderstanding.
So, your original question asks for $$ p=P(X>(n+1)ab) $$
Now, for large $n$, we assume $ X\sim\mathcal{N}(n\mu,n\sigma^2) $ and so we get: $$ p=P\left( \frac{X - nab}{b\sqrt{an}} > \frac{(n+1)ab-nab}{b\sqrt{an}} \right) \approx P\left( Z > \frac{\sqrt{a}}{\sqrt{n}} \right) = 1 - \Phi\left(\sqrt{\frac{a}{n}}\,\right)$$ where $\Phi$ is the normal cdf.
