How to compute the following limit: \begin{align} \lim_{n \to \infty} E \left[ \left(\frac{ \|Z_n\|}{\sqrt{n}+ \sqrt{n +\|Z_n\|^2}} \right)^2 \right] \end{align} where $Z_n=[Z_1,...,Z_n]$ is vector of independent and identical Gaussians each with zero mean and with varieance $\sigma^2$.
At first I tried to use dominated convergence theorem since $\left(\frac{ \|Z_n\|}{\sqrt{n}+ \sqrt{n +\|Z_n\|^2}} \right)^2 \le 1$ and I got a limit of zero. However, I don't think that is correct. What I didn't do is take into account that $Z_n$ also dependes on $n$.
Introduce $$T_n=\frac{ \|Z_n\|}{\sqrt{n}+ \sqrt{n +\|Z_n\|^2}}$$ and note that $$T_n=\frac{Y_n}{1+ \sqrt{1 +Y_n^2}}$$ where $$Y_n=\frac{ \|Z_n\|}{\sqrt n}$$ By the law of large numbers, $$Y_n^2=\frac1n\sum_{k=1}^n(Z_n^{(k)})^2\to\sigma^2\ \text{almost surely}$$ hence $$T_n\to\frac{\sigma}{1+\sqrt{1+\sigma^2}}\ \text{almost surely}$$ Since $0\leqslant T_n\leqslant1$ almost surely, by uniform integrability, this implies that $$\lim_{n \to \infty} E \left[ \left(\frac{ \|Z_n\|}{\sqrt{n}+ \sqrt{n +\|Z_n\|^2}} \right)^2 \right]=\frac{\sigma^2}{(1+\sqrt{1+\sigma^2})^2}$$