Question

$\red{\displaystyle \rm \lim_{n \to \infty }n \int_{0}^{1} \bigg( \frac{2x}{1 + x} \bigg) ^{n} dx}$​

Answer 1

$\red{\displaystyle \rm \lim_{n \to \infty }n \int_{0}^{1} \bigg( \frac{2x}{1 + x} \bigg) ^{n} dx}$

$\red{\displaystyle \rm A_n = \int_{0}^{1} \bigg( \frac{2x}{1 + x} \bigg) ^{n} dx}$

$\large \rm \red{u = \frac{2x}{1 + x} } \: \: \: \: \: \rm \red{ x = \frac{u}{2 - u} }$

$\large \rm \red{dx = \dfrac{1}{2 \left(1 + \frac{u}{2} \right ) {}^{2} }du }$

$\red{\displaystyle \rm \int_{0}^{1} \frac{ {u}^n }{2 \left(1 - \frac{u}{2} \right)^{2} } du = \frac{1}{2}\int_{0}^{1} \frac{ {u}^{n} }{0(1 - \frac{u}{2})(1 - \frac{ u }{2} ) } du }$

$\red{\displaystyle \rm = \frac{1}{2}\int_{0}^{1} {u}^{n} \sum_{k = 0}^{ \infty } \bigg( \frac{u}{2} \bigg)^k \sum_{m = 0}^{ \infty } \bigg( \frac{u}{2} \bigg)^m \: du }$

$\red{\displaystyle \rm = \frac{1}{2}\sum_{k = 0}^{ \infty } \frac{1}{ {2}^{k} } \sum_{m = 0}^{ \infty } \frac{1}{ {2}^{m} } \int_{0}^{1} {u}^{n + k + m} \: du }$

$\red{\displaystyle \rm = \frac{1}{2}\sum_{k = 0}^{ \infty } \frac{1}{ {2}^{k} } \sum_{n = 0}^{ \infty } \frac{1}{ {2}^{m} } \left [ \frac{ {u}^{n + k + m + 1} }{n + k + m + 1} \bigg|_{0}^1 \right] }$

$\red{\displaystyle \rm A_n= \frac{1}{2}\sum_{k = 0}^{ \infty } \frac{1}{ {2}^{k} } \sum_{m = 0}^{ \infty } \frac{1}{ {2}^{m} } \left ( \frac{ 1 }{n + k + m + 1} \right )}$

$\color{red} \displaystyle \rm \lim_{n \to \infty } n \cdot A_n = \lim_{n \to \infty } \frac{1}{2} \sum_{k = 0}^{ \infty } \frac{1}{2m} \overbrace{\bigg( \frac{n}{n + k + m + 1} \bigg)}^{ \displaystyle \rm\lim_{n \to \infty } = 1}$

$\large \rm \red{ \dfrac{1}{2} \bigg( \sum \limits_{k = 0}^ \infty \frac{1}{ {2}^{k} } \bigg)^{2} = \rm \dfrac{1}{2} \bigg( \frac{1}{1 - \frac{1}{2} } \bigg)^{2} }$

$\large \red{ \frac{1}{2} (2) {}^{2} = 2 }$