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

Mathematics
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if there was a pattern we could find from the partial sums, that could be useful; but its still a shot in the dark
Can i use the fact that the sum of 1/k! from 0 to infinity is e-1
i cant say for sure .... you can give it a shot and then verify it with the wolf

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Other answers:

you can try taking the partial sums up to some arbitrary point, and then averaging the integrals of the remainder
but i got no idea how to intgegrate a factorial at the moment
\[e=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\] \[k^2e=k^2\left(\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\right)\] pfft, theres prolly something wrong with my idea on that one
\[\large \frac{ k^2 }{ k! }=\frac{k(k-1+1)}{k!}=\frac{k(k-1)}{k!}+\frac{k}{k!}=\frac{1}{(k-2)!}+\frac{1}{(k-1)!}\]
so the answer is 2e.
good job :)
^^

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