IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Pavis

the sum of series

1+ 2³/2! + 3³/3! ........... is

ans 5e

allamraju

r³/r!=

r²-1+1/(r-1)!=

r+1/(r-2)!+e

₌

r-2+3/(r-2)!+e=e+3e+e=5e

rate me,if i am correct

Pavis

i will rate u first teoll me how {(r-1)! and 1/(r-2)! and 1/(r-3)! }=e

Pavis

first tell how [1/(r-1)! 1/(r-2)! and 1/(r-3)!] =e

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???

krishna.gopal

Perfect answer allamraju. Well done