I have the following mathematical expression:
\begin{align} f(n,m, a)=\frac{1}{n}+\frac{1}{m}+ \frac{a_{n,m}}{(n+m)S_{n,m}} \end{align}
As $n$ and $m$ increases to infinity, $a_{n,m} \overset{p} \rightarrow 0$ or $a$ converges in probability to 0. $S_{n,m}$ converges to a constant term asymptotically. How to formally show $\frac{a_{n,m}}{(n+m)S_{n,m}}$ is lower order term than the first two terms or negligible than the first two terms? The tricky thing is as $n$ and $m$ increases, the first two terms $1/n$ or $1/m$ are also very small.
Visusally, $a=o_p(1)$ and $(n+m)S_{n,m}=O_p(n+m)$. The last term is at lower order term than either $1/n$ or $1/m$.
A more concrete example:
Let's say we have $n_1+n_2$ subjects randomized into two groups . First group has $n_1$ observations $(Y_{1,1},Y_{1,2},\cdots, Y_{1,n_1})$ and Second group has $n_2$ observations $(Y_{2,1},Y_{2,2},\cdots, Y_{2,n_2})$. We would expect asymptotically (as $n_1 \rightarrow \infty$ and $n_2 \rightarrow \infty$), two group means are equal. i.e., $\bar{Y}_{1.}=\bar{Y}_{2.}$ and their group variances are equal. i.e, $S_{n_1}^2=S_{n_2}^2=S^2$
Now I like to show that:
in the following expression:
\begin{align} \frac{1}{n_1}+\frac{1}{n_2}+ \frac{(\bar{Y}_{1.}-\bar{Y}_{2.})^2}{\sum^2_{j=1}\sum^{n_j}_{i=1}(Y_{ij}-\bar{Y}_{j.})^2} \end{align}
the 3rd term is negligible relative to the first two terms or it is at a lower order. As $n_1$ and $n_2$ increases, $(\bar{Y}_{1.}-\bar{Y}_{2.})^2 \overset{p}\rightarrow 0$. $\sum^2_{j=1}\sum^{n_j}_{i=1}(Y_{ij}-\bar{Y}_{j.})^2 \approx n_1 S^2_{n_1}+n_2S^2_{n_2} \approx (n_1+n_2) S^2$.
It is obvious to me that the 3rd term is lower order because numerator $(\bar{Y}_{1.}-\bar{Y}_{2.})^2 \overset{p}\rightarrow 0$ while other two terms have 1, 3 denominators are approximately at the same order. Thus, asymptotically, the 3rd term is negligible relative to the 1st two terms. However, one argument I got is that the other two terms also goes to 0 asymptotically.