Question

How to find the sum of k^2/k! from 0 to infinity?

Answer 1

if there was a pattern we could find from the partial sums, that could be useful; but its still a shot in the dark

Answer 2

Can i use the fact that the sum of 1/k! from 0 to infinity is e-1

Answer 3

i cant say for sure .... you can give it a shot and then verify it with the wolf

Answer 4

you can try taking the partial sums up to some arbitrary point, and then averaging the integrals of the remainder

Answer 5

but i got no idea how to intgegrate a factorial at the moment

Answer 6

\[e=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\]
\[k^2e=k^2\left(\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\right)\]
pfft, theres prolly something wrong with my idea on that one

Answer 7

\[\large \frac{ k^2 }{ k! }=\frac{k(k-1+1)}{k!}=\frac{k(k-1)}{k!}+\frac{k}{k!}=\frac{1}{(k-2)!}+\frac{1}{(k-1)!}\]

Answer 8

so the answer is 2e.

Answer 9

good job :)

Answer 10

^^