I need some quick help on problem 37 from the practice test:
What is the sum of k^2/k! as k = 0 -> infinity?
The booklet claims the answer is 2*e, and I know for a fact that the sum of 1/k! is e, but I can't logically get from one to the other. I've tried looking at Taylor series, but it's no use. Any help would be greatly appreciated.
Problem 37 on the Practice Test - I need help.
The Answer
False alarm, guys. I found the answer on another message board.
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.