Let $T$ be a finite set.
Let $\rho:T\rightarrow (0,1)$ be such that $\sum_{t\in T}\rho(t)=1$.
Let $F:\mathbb N\cup\{0\}\rightarrow(0,1)$ be such that $\sum_{i=0}^\infty F(i)=1$. Let $\mu_F=\sum_{i=0}^\infty iF(i)$.
Let $\ell:T\rightarrow\mathbb N$ be any function.
Fix $t_0\in T$. Then consider the two quantities:
$$\sum_{i=0}^{\infty}F(i)\frac{\ell(t_0)-i}{\sum_{u\in T}\rho(u)\ell(u)-i}$$ and $$\frac{\ell(t_0)-\mu_F}{\sum_{t\in T}\rho(t)\ell(t)-\mu_F}$$
I came across a point in a paper where the second quantity was substituted for the first. But no proof or argument was given as to why this is a reasonable approximation.
So my question is, is there some kind of general theory I can look up that handles approximations such as this? Any guidance would be greatly appreciated.
Thank you!
Abbreviate $$\sum_{i=0}^{\infty}F(i)\frac{\ell(t_0)-i}{\sum_{u\in T}\rho(u)\ell(u)-i}=\sum_{i=0}^{\infty}F_i\frac{a-i}{b-i}$$
We assume that $F_i$ is a decreasing sequence. Note that both $a-i$ and $b-i$ are also decreasing sequences, but that the ratio $\frac{a-i}{b-i}$ can be increasing or decreasing depending on whether $a<b$ or $b<a$, respectively.
Let's assume first that $a<b$. Then, using Chebyshev's sum inequality twice in succession reveals that
$$\begin{align} \sum_{i=0}^{n}F_i\frac{a-i}{b-i} &\le n\left(\frac1n\sum_{i=0}^{n}F_i\right)\left(\frac1n\sum_{i=0}^{n}\frac{a-i}{b-i}\right)\\\\ &=n\left(\frac1n\sum_{i=0}^{n}F_i\right)\left(\frac1n\sum_{i=0}^{n}\frac{F_i(a-i)}{F_i(b-i)}\right)\\\\ & \le n\left(\frac1n\sum_{i=0}^{n}F_i\right)\frac{\frac1n \sum_{i=0}^{n} F_i(a-i)}{\frac1n \sum_{i=0}^{n} F_i(b-i)}\\\\ & \le \left(\sum_{i=0}^{n}F_i\right)\frac{ \sum_{i=0}^{n} F_i(a-i)}{ \sum_{i=0}^{n} F_i(b-i)}\\\\ \end{align}$$
Passing to the limit as $n \to \infty$ gives the desired inequality
$$\begin{align} \lim_{n \to \infty}\sum_{i=0}^{n}F_i\frac{a-i}{b-i} &=\sum_{i=0}^{\infty}F_i\frac{a-i}{b-i}\\\\ &\le \left(\sum_{i=0}^{\infty}F_i\right)\frac{a\sum_{i=0}^{\infty} F_i-\sum_{i=0}^{\infty} F_ii}{b\sum_{i=0}^{\infty} F_i-\sum_{i=0}^{\infty} F_ii}\\\\ &=\frac{a-\mu_F}{b-\mu_F} \end{align}$$
The case for $a>b$ can be analyzed analogously. Obvioulsy, the closer $a$ is to $b$, the tighter the inequality becomes. Recall that $a=\ell(t_0)$ while $b=\sum_{u\in T}\rho(u)\ell(u)$ is an effective averaging of $\ell$ at discrete points.