Question

Show that
$\frac{1}{1-}\frac{1}{3-}\frac{2}{4-}\frac{3}{5-}\cdots \frac{n-1}{n+1-}\cdots =e-1$.

Answer 1

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

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

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

..............................................

$u_3-3u_2=u_2-2u_1$

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

By multiplication;-

$p_1=1,q_1=1,p_2=3,q_2=2,$

$\therefore q_{n+1}-(n+1)q_n=0$

$\therefore q_{n+1}=(n+1)q_n=(n+1)n{q}_{n-1}$

$=......=(n+1)!$

Again,

$p_{n+1}-(n+1)p_n=1$

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

$\frac{p_2}{2!}-\frac{p_1}{1!}=\frac{1}{2!}$

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

$\therefore \frac{p_{n+1}}{q_{n+1}}=1+\frac{1}{2!}+\frac{1}{2!}+.....+\frac{1}{(n+1)!}$

$\therefore \frac{p_{\infty }}{q_{\infty }}=e-1$