Let $X_1, X_2, \dots , X_m$ and $Y_1, Y_2, \dots , Y_n$ be two independent random samples with means and variances equal to $(\mu _1, \sigma _1 ^2)$ and $(\mu _2, \sigma _2 ^2)$, respectively. Let $T_{m,n} = (\overline{X}_m - \overline{Y} _n) / S_{m,n}$ with $S_{m,n} ^2 = S_X ^2 / m + S_Y ^2 /n$, where $S_X ^2$ and $S_Y ^2$ are the sample variances of the two samples.
I am doing an exercise (not for a grade) in which I have to show that the test that rejects $H_0 : \mu _1 = \mu_2$ if $|T_{m,n}| > \xi _{\alpha}$ is of asymptotic level $2\alpha$.
I have already proven this result for $m=n\to \infty$.
Write $T_{m,n} = \frac{\sqrt{n} ( \overline{X} _n - \overline{Y} _n ) }{\sqrt{S_X ^2 + S_Y ^2}}$. Notice that $S_X ^2 \overset{P}{\longrightarrow} \sigma _1 ^2$ and $S_Y ^2 \overset{P}{\longrightarrow} \sigma _2 ^2$. By the continuous mapping theorem $\sqrt{S_X ^2 + S_Y ^2} \overset{P}{\longrightarrow} \sqrt{\sigma _1 ^2 + \sigma _2 ^2 }$. Lets assume the null hypothesis is true. By the central limit theorem: $\sqrt{n} (\overline{X} _n - \mu _1 ) \overset{d}{\longrightarrow} \mathcal{N} (0, \sigma _1 ^2)$ and $\sqrt{n} (\overline{Y} _n - \mu _1 ) \overset{d}{\longrightarrow} \mathcal{N} (0, \sigma _2 ^2)$. Thus, by a result about the weak convergence of weakly converging independent sequences, and by continuous mapping:$\sqrt{n} (\overline{X} _n - \overline{Y} _n ) \overset{d}{\longrightarrow} \mathcal{N} (0, \sigma _1 ^2 + \sigma _2 ^2 )$. Now, by Slutsky $T_{m,n} = \frac{\sqrt{n} ( \overline{X} _n - \overline{Y} _n ) }{\sqrt{S_X ^2 + S_Y ^2}} \overset{d}{\longrightarrow} \mathcal{N} (0,1)$.
How can I use the result for $m=n\to \infty$ in order to prove the result for general $m,n\to \infty$?
Let $m_k$ and $n_k$ denote the corresponding sample sizes s.t. $m_k\to\infty$, $n_k\to \infty$, and $$ \frac{a_k}{\sqrt{m_k}}\to a_1<\infty\quad\text{and}\quad \frac{a_k}{\sqrt{n_k}}\to a_2<\infty $$ as $k\to\infty$, where $\{a_k\}$ is a sequence of positive numbers. Then as in the answer to this post, $$ a_k\!\left((\bar{X}_{m_k}-\bar{Y}_{n_k})-(\mu_1-\mu_2)\right)\xrightarrow{d}\mathcal{N}\!\left(0,a_1^2\sigma_1^2+a_2^2\sigma_2^2\right). $$ Aslo $$ a_k^2\left(\frac{S_X^2}{m_k}+\frac{S_Y^2}{n_k}\right)\xrightarrow{p}a_1^2\sigma_1^2+a_2^2\sigma_2^2. $$ Assuming that $a_1\vee a_2>0$, $T_{m_k,n_k}\xrightarrow{d}\mathcal{N}(0,1)$.