Limit $\lim_{n\to\infty}\sum_{k=0}^n\binom nk\frac{3k}{2^n(n+3k)}$ 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.
 A: 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 $\frac35$.
Similar question: Proof/derivation of $\lim\limits_{n\to\infty}{\frac1{2^n}\sum\limits_{k=0}^n\binom{n}{k}\frac{an+bk}{cn+dk}}\stackrel?=\frac{2a+b}{2c+d}$?
A: Let $(s_n)_{n\in\mathbb{N}_0}$ be defined as $s_n=\sum_{k=0}^{n}\binom{n}{k}\frac{3k}{2^n(n+3k)}$. We're going to show that $\lim_{n\to\infty}s_n=\frac{3}{5}$.
First some "moving things around":
$$
s_n=\sum_{k=0}^{n}\binom{n}{k}\frac{3k}{2^n(n+3k)}=\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2^n}\left(1-\frac{n}{n+3k}\right)=\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2^n}-\sum_{k=0}^{n}\binom{n}{k}\frac{1}{2^n}\frac{n}{n+3k}\\=
1-\frac{n}{2^n}\sum_{k=0}^{n}\binom{n}{k}\frac{1}{n+3k}
$$
Thus, lets consider $t_n=1-s_n$:
$$
t_n=\frac{n}{2^n}\sum_{k=0}^{n}\binom{n}{k}\frac{1}{n+3k}=\frac{n}{2^n}\sum_{k=0}^{n}\left(\binom{n}{k}\int_0^1 x^{n+3k-1}dx\right)=\frac{n}{2^n}\int_0^1 \left(x^{n-1}\sum_{k=0}^{n}\binom{n}{k}x^{3k}\right)dx=\frac{n}{2^n}\int_0^1 x^{n-1}\left(1+x^3\right)^ndx=\frac{n}{2^n}\int_0^1 \frac{1}{n}y^{\frac{1}{n}-1}y^{1-\frac1n}\left(1+y^{\frac{3}{n}}\right)^ndy=\int_0^1 \left(\frac{1+y^{\frac{3}{n}}}{2}\right)^ndy
$$
Where in the last integral we used a change of variable $x=y^{\frac{1}{n}}$. Now let $f_n(y):=\left(\frac{1+y^{\frac{3}{n}}}{2}\right)^n$. For $y\in\mathbb{R}_{≥0}$ fixed we see:
$$
\log\left(\lim_{n\to\infty}f_n(y)\right)=\lim_{n\to\infty}\log\left(f_n(y)\right)=\lim_{n\to\infty}n\log\left(\frac{1+y^{\frac{3}{n}}}{2}\right)=\lim_{n\to\infty}\frac{\log\left(\frac{1+y^{\frac{3}{n}}}{2}\right)}{\frac{1}{n}}=\lim_{n\to\infty}\frac{-\frac{1}{n^2}\log(y^3)\left(\frac{y^{\frac{3}{n}}}{1+y^{\frac{3}{n}}}\right)}{-\frac{1}{n^2}}=3\log(y)\lim_{n\to\infty}\frac{y^{\frac{3}{n}}}{1+y^{\frac{3}{n}}}=\frac{3}{2}\log(y)\implies \lim_{n\to\infty}f_n(y)=y^{\frac{3}{2}}=:f(y)
$$
Where we used L'Hopitals rule and the fact that $\lim_{n\to\infty}y^{3/n}=1$. Furthermore, one can prove using Hölder (see below)
$$
f_n(y)\ge f_{n+1}(y)
$$
for all $y\in\mathbb{R}_{≥0}$ and $n\in\mathbb{N}_{0}$. Together with Dinis theorem, this shows that that $f_n\rightarrow f$ uniformly on any intervall $[0,R]$. Thus:
$$
\lim_{n\to\infty}t_n=\lim_{n\to\infty}\int_0^1 f_n(y)dy=\int_0^1 \lim_{n\to\infty}f_n(y)dy=\int_0^1 y^{\frac{3}{2}}dy=\frac{2}{5}
$$
Which proves that:
$$
\lim_{n\to\infty}\sum_{k=0}^{n}\binom{n}{k}\frac{3k}{2^n(n+3k)}=\lim_{n\to\infty} s_n = 1-\frac{2}{5}=\frac{3}{5}
$$
Edit:
Here is a way to prove $f_n(y)≥f_{n+1}(y)$ for all $y\in\mathbb{R}_{≥0}$ and $n\in\mathbb{N}$ which only uses the Hölder inequality:
$$
f_n(y)≥f_{n+1}(y) \iff \left(\frac{1+y^{\frac{3}{n}}}{2}\right)^n≥\left(\frac{1+y^{\frac{3}{n+1}}}{2}\right)^{n+1}\iff 2\left(1+y^{\frac{3}{n}}\right)^n≥\left(1+y^{\frac{3}{n+1}}\right)^{n+1}
$$
This last inequality is true because by Hölder
$$
2\left(1+y^{\frac{3}{n}}\right)^n=(1+1)\underbrace{\left(1+y^{\frac{3}{n}}\right)\cdots \left(1+y^{\frac{3}{n}}\right)}_{n\text{ times}}≥\left(1+\left(1\cdot\underbrace{y^{\frac{3}{n}}\cdots y^{\frac{3}{n}}}_{n\text{ times}}\right)^{\frac{1}{n+1}}\right)^{n+1}=\left(1+y^{\frac{3}{n+1}}\right)^{n+1}
$$
