Question

Find sum of the series

$\frac{1}{3.1!}+\frac{1}{4.2!}+\frac{1}{5.3!}+....\infty =$

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. $3$

Answer 1

The correct option is C $\frac{1}{2}$

From the given series, we can write,

$T_n=\frac{1}{(n+2)(n!)}=\frac{(n+1)}{(n+2)(n+1)(n!)}$
$=$ $\frac{(n+1)}{(n+2)!}=\frac{n+2}{(n+2)!}-\frac{1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$
So,
$S=$ $(\frac{1}{2!}-\frac{1}{3!})+(\frac{1}{3!}-\frac{1}{4!})+.........+.....$

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

Cancelling out the terms we get,

$S=\frac{1}{2}$