How can one proof the following:
Let $S_n$ be $\chi_n^2$-distributed. Then $\sqrt{S_n}-\sqrt{n}$ is asymptotically $N(0,\frac{1}{2})$ distributed, i.e. $\sqrt{S_n}-\sqrt{n}$ ~. $N(0,\frac{1}{2})$.
I wanted to use the follwing result: If $\sqrt{n}(T_n-n)$ ~. $N(0, \sigma^2)$, then for a differentiable function $h$ with $h(\mu) \neq 0$ we get
$\sqrt{n}(h(T_n)-h(\mu))$ ~. $N(0,\sigma^2 h'(\mu)^2),$
but I don't know how to start. Can someone give me hint or help me?
We have $$f_{S_n}(x) = \frac{x^{n/2-1} e^{-x/2}}{2^{n/2} \Gamma(n/2)}, \quad x > 0.$$ Thus, $Y = \sqrt{S_n} - \sqrt{n}$ has density $$f_Y(y) = f_{S_n}((y+\sqrt{n})^2) 2(y+\sqrt{n}) = \frac{(y+\sqrt{n})^{n-1} e^{-(y+\sqrt{n})^2/2}}{2^{n/2-1} \Gamma(n/2)}, \quad y > -\sqrt{n}.$$ Using Stirling's approximation for $$\Gamma(n/2) \sim \sqrt{\pi} \sqrt{n - 1/2} \left(\frac{n/2-1}{e}\right)^{n/2-1} + O(1/n)$$ and taking the logarithm, we get for $n > 2$ $$\log (\sqrt{\pi} f_Y(y)) \approx \frac{n-1}{2} \log \frac{(y + \sqrt{n})^2}{n-2} - \frac{y^2}{2} - y \sqrt{n} -1,$$ the limit of which as $n \to \infty$ is $-y^2$, by a straightforward calculation (e.g. L'Hopital's rule). Therefore, as $n \to \infty$ we have $$f_Y(y) \to e^{-y^2}/\sqrt{\pi},$$ which is normal with mean $0$ and variance $1/2$.