Consider a random sample of size $n_{1}$ from the $N(\mu_{1},\sigma_{1}^{2})$ distribution and an independent random sample of size $n_{2}$ from the $N(\mu_{2},\sigma_{2}^{2})$ distribution. Let $S_{1}^{2}$ and $S_{2}^{2}$ denote the sample variances for the two respective samples.
Show that $$ \sqrt{\frac{S_{1}^{2}}{n_{1}} + \frac{S_{2}^{2}}{n_{2}}} \Bigg/ \sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}} \stackrel{P}{\to} 1. $$
I am a bit confused as to where to start here any tips? Maybe somehow show the top converges in probability to the bottom and the bottom converges to itself and apply Slutsky's theorem?
Remove square root first. Next introduce two sequences $$ \alpha_{n_1} = S_1^2 - \sigma_1^2 \stackrel{P}{\to} 0 \text{ as } n_1\to\infty $$ and $$ \beta_{n_2} = S_2^2 - \sigma_2^2 \stackrel{P}{\to} 0 \text{ as } n_2\to\infty . $$ Replace $S_1^2$ by $\sigma_1^2+\alpha_{n_1}$ and $S_2^2$ by $\sigma_2^2+\beta_{n_2}$ in the initial fraction and get $$ \frac{\frac{S_{1}^{2}}{n_{1}} + \frac{S_{2}^{2}}{n_{2}}}{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}} = 1+ \frac{\alpha_{n_1}}{\sigma_1^2+\frac{n_1}{n_2}\sigma_2^2}+\frac{\beta_{n_2}}{\sigma_2^2+\frac{n_2}{n_1}\sigma_1^2} $$ Next, $$ \frac{|\alpha_{n_1}|}{\sigma_1^2+\frac{n_1}{n_2}\sigma_2^2}\leq \frac{|\alpha_{n_1}|}{\sigma_1^2} \stackrel{P}{\to} 0 \text{ as } n_1\to\infty, $$ and the last summand behaves the same way.