Proof/derivation of $\lim\limits_{n\to\infty}{\frac1{2^n}\sum\limits_{k=0}^n\binom{n}{k}\frac{an+bk}{cn+dk}}\stackrel?=\frac{2a+b}{2c+d}$?

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.

The reason this is true is that the binomial coefficients are strongly concentrated around the mean, when $k \sim \frac{n}{2}$. Using some standard concentration inequalities (Chernoff is strong, but Chebyshev's inequality sufficies too), you can show that for any constant $c > 0$, $2^{-n} \sum_{ k \in [0, (1/2 - c ) n ] \cup [(1/2 + c)n, n]} \binom{n}{k} \rightarrow 0$ as $n \rightarrow \infty$.

Hence in your limit, the only terms that survive are when $k \sim \frac{n}{2}$, in which case you can cancel the $n$ throughout and get the right-hand side.

For the limit to hold, you clearly need $2c + d \neq 0$. You would also need $c \neq 0$, as otherwise the $k = 0$ term of the sum will present difficulties.

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\lim_{n\to \infty}\,{1 \over 2^{n}} \sum_{k = 0}^{n}{n \choose k}{an + bk \over cn + dk}} = \lim_{n\to \infty}\,{1 \over 2^{n}} \sum_{k = 0}^{n}{n \choose k}\pars{an + bk}\int_{0}^{1}t^{cn + dk - 1}\,\,\,\dd t \\[3mm] = &\ \lim_{n\to \infty}\,{1 \over 2^{n}}\int_{0}^{1}t^{cn - 1}\bracks{% an\sum_{k = 0}^{n}{n \choose k}\pars{t^{d}}^{k} + b\sum_{k = 0}^{n}{n \choose k}k\pars{t^{d}}^{k}}\dd t \end{align}

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However, $\ds{\sum_{k = 0}^{n}{n \choose k}\xi^{k} = \pars{1 + \xi}^{n} \quad\imp\quad \sum_{k = 0}^{n}{n \choose k}k\,\xi^{k} = n\xi\pars{1 + \xi}^{n - 1}}$.

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Then, \begin{align} &\color{#f00}{\lim_{n\to \infty}\,{1 \over 2^{n}} \sum_{k = 0}^{n}{n \choose k}{an + bk \over cn + dk}} = \\[3mm] = &\ \lim_{n\to \infty}\,{1 \over 2^{n}}\int_{0}^{1}t^{cn - 1}\bracks{% an\pars{1 + t^{d}}^{n} + bn\,t^{d}\pars{1 + t^{d}}^{n - 1}}\dd t \\[3mm] = &\ \color{#f00}{\lim_{n \to \infty}\braces{\vphantom{\LARGE A}% {n \over 2^{n}}\bracks{\vphantom{\Large A}a\,\mathrm{f}_{n}\pars{cn,d} + b\,\mathrm{f}_{n - 1}\pars{cn + d,d}}}} \end{align}

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$$ \mbox{where}\quad \begin{array}{|c|}\hline\\ \ds{\quad\mathrm{f}_{n}\pars{\mu,\nu} \equiv \int_{0}^{1}t^{\mu - 1}\pars{1 + t^{\nu}}^{n}\,\dd t\quad} \\ \\ \hline \end{array} $$

Can you take it from here ?. Maybe, an asymptotic study of the integral could be somehow reasonable. Otherwise, the integral is related to hypergeometric functions.

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.