anonymous
  • anonymous
How to find the sum of k^2/k! from 0 to infinity?
Mathematics
katieb
  • katieb
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

amistre64
  • amistre64
if there was a pattern we could find from the partial sums, that could be useful; but its still a shot in the dark
anonymous
  • anonymous
Can i use the fact that the sum of 1/k! from 0 to infinity is e-1
amistre64
  • amistre64
i cant say for sure .... you can give it a shot and then verify it with the wolf

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
you can try taking the partial sums up to some arbitrary point, and then averaging the integrals of the remainder
amistre64
  • amistre64
but i got no idea how to intgegrate a factorial at the moment
amistre64
  • amistre64
\[e=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\] \[k^2e=k^2\left(\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\right)\] pfft, theres prolly something wrong with my idea on that one
sirm3d
  • sirm3d
\[\large \frac{ k^2 }{ k! }=\frac{k(k-1+1)}{k!}=\frac{k(k-1)}{k!}+\frac{k}{k!}=\frac{1}{(k-2)!}+\frac{1}{(k-1)!}\]
sirm3d
  • sirm3d
so the answer is 2e.
amistre64
  • amistre64
good job :)
sirm3d
  • sirm3d
^^

Looking for something else?

Not the answer you are looking for? Search for more explanations.