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.
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1$$(n+1)I_n=\int_0^1\frac{du}{1+u^{1/(n+1)}}\to\int_0^1\frac{du}{1+1}$$ – Did Apr 22 '16 at 09:14
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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
$$ \lim_{n \to +\infty}nI_n=\frac12.$$
Olivier Oloa
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Pour les inégalités (2) quel(le) théorème ou propriété vous permet de justifier la deuxième inégalité? – ParaH2 Apr 22 '16 at 09:27
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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 09:30
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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 $$ \lim_{n\to \infty}(n+1)I_{n} = \dfrac{1}{2}. $$
JanG
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