Limit of sum with binomial coeffs Let
$$S_{n}=\frac{1}{2^n}\sum_{k=1}^{n}{n\choose k}\frac{k}{n+k}$$,
find the
$$y=\lim_{n\to +\infty}{S_{n}}$$
I can prove that $ \frac{1}{4} \leq S_{n} \leq \frac{1}{2}$:
$$ {n\choose k}\frac{k}{2n} \leq {n\choose k}\frac{k}{n+k} \leq {n\choose k}\frac{k}{n} $$
$$ \frac{1}{2}\sum_{k=1}^{n}{n\choose k}\frac{k}{n} \leq \sum_{k=1}^{n}{n\choose k}\frac{k}{n+k} \leq \sum_{k=1}^{n}{n\choose k}\frac{k}{n} $$
$$ \frac{1}{2}\sum_{k=1}^{n}{n-1\choose k-1} \leq \sum_{k=1}^{n}{n\choose k}\frac{k}{n+k} \leq \sum_{k=1}^{n}{n-1\choose k-1} $$
$$2^{n-2} \leq \sum_{k=1}^{n}{n\choose k}\frac{k}{n+k} \leq 2^{n-1} $$
$$\frac{1}{4} \leq \frac{1}{2^n}\sum_{k=1}^{n}{n\choose k}\frac{k}{n+k} \leq \frac{1}{2} $$
Numeric evaluation shows that probably $y=\frac{1}{3}$.
Also, I tried to apply differentiation and integration:
$$(1+x)^n=\sum_{k=0}^{n}{n\choose k}{x^k}$$
differentiate:
$$n(1+x)^{n-1}=\sum_{k=1}^{n}{n\choose k}k{x^{k-1}}$$
multiply by $x^n$:
$$n{x^n}(1+x)^{n-1}=\sum_{k=1}^{n}{n\choose k}k{x^{n+k-1}}$$
take the $\int_{0}^{1}$:
$$n\int_{0}^{1}{x^n}(1+x)^{n-1}\,dx=\sum_{k=1}^{n}{n\choose k}\frac{k}{n+k}$$
that finally gives:
$$S_n=\frac{n}{2^n}\int_{0}^{1}{x^n}{(1+x)^{n-1}}\,dx$$
But I stuck finding the limit of this expression as ${n\to +\infty}$
 A: Since we can transform the integral into $$\frac12\int_{0}^{1}x\cdot n\left[\frac{x(1+x)}{2}\right]^{n-1}\,dx,$$ it's natural to substitute $$t=\frac{x(1+x)}{2},$$ i.e. $$x=\sqrt{2t+\frac14}-\frac12.$$ Then, we have
$$S_n=\int_{0}^{1}\,\frac{\sqrt{2t+\frac14}-\frac12}{2\sqrt{2t+\frac14}}\cdot nt^{n-1}dt=\int_{0}^{1}f(t)\cdot nt^{n-1}dt.$$ Since $f$ has no singularities in $[0,1]$, we can integrate by parts:
$$S_n=f(1)-\int_{0}^{1}f'(t)\,t^n\,dt.$$ But $f(1)=\frac13$, and the absolute value of the second integral is bounded by $\int_{0}^{1}C\,t^n\,dt=\frac{C}{n+1}$. So we have $$\lim_{t\to\infty}S_n=\frac13.$$  
A: This has a simple probabilistic interpretation. 
Namely, 
$$
S_n = \mathrm{E}\Big[\frac{Z_n}{n+Z_n}\Big] = \mathrm{E}\Big[\frac{\frac1nZ_n}{1+\frac1n Z_n}\Big],
$$
where $Z_n$ has a binomial $(n,\frac12)$ distribution. We know that $Z_n \overset{d}= X_1 + \dots + X_n$, where $X_k$ are iid Bernoulli$(\frac12)$. By LLN, $\frac 1n Z_n\overset{\mathrm{P}}{\longrightarrow} \frac12$, $n\to\infty$. Therefore, in view of the dominated convergence,
$$
S_n \to \frac{\frac12}{1+\frac12} = \frac{1}{3}.
$$
