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Rachel123

  • 2 years ago

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

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  1. amistre64
    • 2 years ago
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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

  2. Rachel123
    • 2 years ago
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    Can i use the fact that the sum of 1/k! from 0 to infinity is e-1

  3. amistre64
    • 2 years ago
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    i cant say for sure .... you can give it a shot and then verify it with the wolf

  4. amistre64
    • 2 years ago
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    you can try taking the partial sums up to some arbitrary point, and then averaging the integrals of the remainder

  5. amistre64
    • 2 years ago
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    but i got no idea how to intgegrate a factorial at the moment

  6. amistre64
    • 2 years ago
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    \[e=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\] \[k^2e=k^2\left(\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\frac1{4!}+\frac1{5!}+...\right)\] pfft, theres prolly something wrong with my idea on that one

  7. sirm3d
    • 2 years ago
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    \[\large \frac{ k^2 }{ k! }=\frac{k(k-1+1)}{k!}=\frac{k(k-1)}{k!}+\frac{k}{k!}=\frac{1}{(k-2)!}+\frac{1}{(k-1)!}\]

  8. sirm3d
    • 2 years ago
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    so the answer is 2e.

  9. amistre64
    • 2 years ago
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    good job :)

  10. sirm3d
    • 2 years ago
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    ^^

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