TITLE: Proof/derivation of $\lim\limits_{n\to\infty}{\frac1{2^n}\sum\limits_{k=0}^n\binom{n}{k}\frac{an+bk}{cn+dk}}\stackrel?=\frac{2a+b}{2c+d}$?
QUESTION [30 upvotes]: I just came up with the following identity while solving some combinatorial problem but not sure if it's correct. I've done some numerical computations and they coincide.
$$\lim_{n\to \infty}{\frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}\frac{an+bk}{cn+dk}}\;\stackrel?=\;\frac{2a+b}{2c+d}$$
Here $a$, $b$, $c$, and $d$ are reals except that $c$ mustn't $0$ and $2c+d\neq0$. I wish I could explain how I came up with it, but I did nothing but comparing numbers with the answer then formulated the identity, and just did numerical computations.

REPLY [4 votes]: I believe a probabilistic argument can be given here.
Let $X_1, X_2, \ldots $ be iid Bernoulli random variables with success probability $p$. Then by Strong Law of Large Numbers $$\bar{X}_n:=\frac{1}n\sum_{i=1}^n X_i \stackrel{a.s.}{\to} p$$
Also for any continuous function $g$ we have $g(\bar{X}_n) \stackrel{a.s.}{\to} g(p)$. Now if $g$ is such that there exist a random variable $Y$ with finite expectation such that $g(\bar{X}_n) \le Y$ almost surely. Then by DCT we have $E(g(\bar{X}_n)) \to g(p)$.
Now this has lot of applications.
Example 1: Take $p=\frac12$. $g(x)=\frac{a+bx}{c+dx}$. Make sure $g$ is continuous $[0,1]$. Then clearly $g$ is bounded. Hence by DCT
$$E(g(\bar{X}_n))=\sum_{i=1}^n \frac{\binom{n}{k}}{2^n}g\left(\frac{k}{n}\right) \to g(\frac12)=\frac{2a+b}{2c+d}$$
Example 2: Take $p=\frac12$,$g(x)=e^{-x^2}$. Clearly $g$ is continuous and bounded.
Hence
$$\frac1{2^n}\sum_{i=1}^n \binom{n}{k}e^{-k^2/n^2} \to e^{-1/4}$$
Similarly we can obtain plethora of such results.