Question

Show that
$\frac{4}{1+}\frac{6}{2+}\frac{8}{3+}\cdots \frac{2n+2}{n+}\cdots =\frac{2(e^2-1)}{e^2+1}$.

Answer 1

$\Rightarrow u_n=n{u}_{n-1}+(2n+2)u_{n-2}$

$\therefore u_n-(n+2)u_{n-1}=-2\{u_{n-1}-(n+1\left)u_{n-2}\right\}$

By multiplication

$u_3-5u_2=-2(u_2-4u_1)$

$u_n-(n+2)u_{n-1}={(-1)}^{n-2}{2}^{n-2}(u_2-4u_1);$

$p_1=4,q_1=1,p_2=8,q_2=8,$

$\therefore p_n-(n+2)p_{n-1}={(-1)}^{n-1}{2}^{n+1},q_n-(n+2)q_{n-1}={(-1)}^{n-1}{2}^n$

$\frac{p_n}{(n+2)!}-\frac{p_{n-1}}{(n+1)!}=\frac{{(-1)}^{n-1}{2}^{n+1}}{(n+2)!}$

$\frac{p_2}{4!}-\frac{p_1}{3!}=\frac{(-1)2^3}{4!}$

$\frac{p_n}{(n+2)!}=\frac{2^2}{3!}-\frac{2^3}{4!}+\frac{2^4}{5!}-.......$

$\frac{q_n}{(n+2)!}-\frac{q_{n-1}}{(n+1)!}=\frac{{(-1)}^{n-1}{2}^n}{(n+2)!}$

$\therefore \frac{q_n}{(n+2)!}=\frac{1}{3!}+\frac{2^2}{4!}-\frac{2^3}{5!}+\frac{2^4}{6!}-.....$

$\therefore \frac{p_n}{q_n}=\frac{1}{2}(\frac{2^3}{3!}-\frac{2^4}{4!}+\frac{2^5}{5!}....)÷\frac{1}{4}(\frac{4}{3!}+\frac{2^4}{4!}-\frac{2^5}{5!}+....)$

$e^{-1}=1-2+\frac{2^2}{2!}-\frac{2^3}{3!}+\frac{2^4}{4!}-...$

$\frac{p_{\infty }}{q_{\infty }}=\frac{1}{2}(1-2+\frac{2^2}{2!}-e^{-2})÷\frac{1}{4}(\frac{4}{3!}+e^{-2}-1+2-\frac{2^2}{2!}+\frac{2^3}{3!})$

$\frac{p_{\infty }}{q_{\infty }}=\frac{1}{2}(1-e^{-2})÷\frac{1}{4}(1+e^{-2})$

$=\frac{2(1-e^{-2})}{1+e^{-2}}=\frac{2(e^2-1)}{e^2+1}$