Prove that $(n+1)\int_0^1{\frac{x^n}{x+1}dx}\to\frac12$ [duplicate]

Let $I_n=\int_0^1{\frac{x^n}{x+1}dx}$ , $n>0$

Show that $\lim_{n->\infty}{(n+1)I_n} = \frac{1}{2}$

All I could do was to show that the $I_n$ is decreasing.

• $$(n+1)I_n=\int_0^1\frac{du}{1+u^{1/(n+1)}}\to\int_0^1\frac{du}{1+1}$$ – Did Apr 22 '16 at 9:14

You may just integrate by parts, \begin{align} I_n=\int_0^1\frac{x^n}{x+1}dx&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{(x+1)}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\\\ &=\frac1{2(n+1)}+\color{blue}{\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx} \tag1 \end{align} then observe that $$0\leq \color{blue}{\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx}\leq \frac{1}{(n+1)}\frac{1}{(0+1)^2}\int_0^1x^{n+1}dx=\frac{1}{(n+1)(n+2)}\tag2$$ Then using $(1)$ and $(2)$ gives easily
$$\lim_{n \to +\infty}nI_n=\frac12.$$
• Le simple fait que $\frac1{(x+1)^2}\leq \frac{1}{(0+1)^2}$ sur l'intervalle $[0,1]$. – Olivier Oloa Apr 22 '16 at 9:30
Since $$\dfrac{x^n}{1+1}\le \dfrac{x^n}{x+1} \le \dfrac{x^n}{x+x}$$ we get that $$\dfrac{1}{2(n+1)} \le I_{n} \le \dfrac{1}{2n}.$$ Consequently $$\lim_{n\to \infty}(n+1)I_{n} = \dfrac{1}{2}.$$