TITLE: Limit $\lim_{n\to\infty}\sum_{k=0}^n\binom nk\frac{3k}{2^n(n+3k)}$
QUESTION [6 upvotes]: Is there a closed-form solution for this combinatorial limit of a sum?
$$\lim_{n\to\infty}\sum_{k=0}^n\binom nk\frac{3k}{2^n(n+3k)}$$
I tried hypergeometric series and failed.

REPLY [8 votes]: Let $X_1, X_2, \ldots $ be iid Bernoulli random variables with success probability $1/2$. Then by Strong Law of Large Numbers $$\bar{X}_n:=\frac{1}n\sum_{i=1}^n X_i \stackrel{a.s.}{\to} \frac12$$
Also for any continuous function $g$ we have $g(\bar{X}_n) \stackrel{a.s.}{\to} g(1/2)$. Now take $g(x)=\frac{3x}{1+3x}$. $|g(\bar{X}_n)| \le 1$ for all $n$. Then by DCT we have $$E(g(\bar{X}_n)) \to g(1/2)=\frac35$$
Let $Y=n\bar{X}_n$. $Y\sim \operatorname{Bin}(n,1/2)$. Finally note that
$$E(g(\bar{X}_n))=E\left(\frac{3\bar{X}_n}{1+3\bar{X}_n}\right)=E\left(\frac{3Y}{n+3Y}\right)=\frac1{2^n}\sum_{i=1}^n \binom{n}{k}\frac{3k}{n+3k}$$
This shows that the limit is indeed $\frac35$.
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