I have the following problem,
Show $1 + z_{\alpha/2}/\sqrt{n} \approx \chi^2_{\alpha/2} (2n)/(2n)$
where $z_{\alpha/2}$ is $\alpha/2$ upper quantile of $N(0,1)$, and $\chi^2_{\alpha/2}(2n)$ is $\alpha/2$ upper quantile of $\chi^2(2n)$ where $2n$ denotes the degree of freedom.
Try
By CLT, we have $\frac{X - 2n}{\sqrt{4n}} \to N(0,1)$, where $X \sim \chi^2(2n)$ since $\chi^2(2n) \equiv \sum_i^n Z_i$ where $Z_i \overset{iid}{\sim} \Gamma(1,2)$. But I cannot proceed.
It seems I found it.
By CLT $\frac{\chi^2_{\alpha/2} (2n) - 2n}{\sqrt{4n}} \approx z_{\alpha/2}$.
By rearranging, $z_{\alpha/2} \approx \sqrt{n}(\chi^2_{\alpha/2} (2n) - 2n)/(2n)$.
thus we have $1 + z_{\alpha/2}/\sqrt{n} \approx \chi^2_{\alpha/2}(2n)$