Limit of sum with binomial coeffs

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 $$\lim_{t\to\infty}S_n=\frac13.$$

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, $$ S_n \to \frac{\frac12}{1+\frac12} = \frac{1}{3}. $$