Dear all,
Can anyone give me some idea on the infinite serious below?
{K=1 to infinity}, (K^2) / K! = ??
Thanks
9768 #37
-
- Posts: 34
- Joined: Thu Dec 30, 2010 4:36 am
Re: 9768 #37
Please note that the guys have previously answered it, as the following :
"
First note that, for k >= 1, k^2/k! = k/(k-1)!. By a change of
variable we may write sum(k=1 to oo) k/(k-1)! = sum(k=0 to oo) (k+1)/
k! = sum(k=0 to oo) 1/k! + sum(k=0 to oo) k/k!. The first term is e.
For the second term note that for the case k=0 the summand is zero, so
we can write this as sum(k=1 to oo) k/k! = sum(k=1 to oo) 1/(k-1)! =
sum(k=0 to oo) 1/k!, where the last equality follows from another
change of variable. So this term is also e."
"
First note that, for k >= 1, k^2/k! = k/(k-1)!. By a change of
variable we may write sum(k=1 to oo) k/(k-1)! = sum(k=0 to oo) (k+1)/
k! = sum(k=0 to oo) 1/k! + sum(k=0 to oo) k/k!. The first term is e.
For the second term note that for the case k=0 the summand is zero, so
we can write this as sum(k=1 to oo) k/k! = sum(k=1 to oo) 1/(k-1)! =
sum(k=0 to oo) 1/k!, where the last equality follows from another
change of variable. So this term is also e."