Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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 :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question