IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ramkumar_november

$\sum_{r=1}^{\infty}\frac{1}{(r-3)!}=\sum_{r=1}^{\infty}\frac{(r-1)(r-2)}{(r-1)!}$

= $0+0+1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+.........$

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

= e

$\sum_{r=1}^{\infty}\frac{1}{(r-2)!}=\sum_{r=1}^{\infty}\frac{(r-1)}{(r-1)!}$

= $0+\frac{1}{1!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+......$

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

= e

$\sum_{r=1}^{\infty}\frac{1}{(r-1)!}=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....$ = e

pavis ..... you clear???