Evaluate this series: (1 Viewer)

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Evaluate this series: Sum from k = 1 to k = infinity k^2/k!

The answer is 2e.

Thanks so much!!
 

Hurkyl

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What have you tried so far?
 
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I have no idea how to do this problem. Does it require more knowledge than second year college math? I thought it just requires some knowledge about series, but I can't seem to get the answer. It's been over 3 years since I took calculus so I don't remember much about series expansions. I've tried comparing the series to several common expansions but I can't find the right form or find the right starting point. Thanks!
 
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yxgao,
here's a hint:
[tex]
e = \sum_{k=0}^\infty \frac{1}{k!}
[/tex]
Try to express your sum in terms of this...
 

Hurkyl

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Have you ever heard of a trick called "integrating with respect to 1"?


Anyways, one of the first things I'd do is simplify the summand; something cancels.
 

krab

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Here's a hint: replace the index k by k+1. You'll be able to cancel k+1 in numerator and denominator. The rest is easy.
 
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How do you simplify (n+1)/n! summed from n = 0 to n = infinity?
 

Hurkyl

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distribute!
 
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I don't understand how to proceed from this step. What do you do with the (n+1) in the numerator? What is there to distribute? Can someone post a direct method to the solution?
Thanks.
 

Hurkyl

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[tex]
\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}
[/tex]
 
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Ah, ok. Why didn't I see that before? :)
e + e = 2e
Thanks a lot.
 

Hurkyl

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It's a slick trick, I haven't seen this particular sum done this way before. The method I know can be done as follows (it's more complicated, but more powerful):


The taylor series for [itex]e^x[/itex] is:

[tex]e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}[/tex]

I can insert a [itex]k[/itex] term by differentiating:

[tex]
e^x = \sum_{k=0}^{\infty} \frac{k x^{k-1}}{k!}
[/tex]

I can then multiply by [itex]x[/itex]:

[tex]x e^x = \sum_{k=0}^{\infty} \frac{k x^{k}}{k!}[/tex]

and differentiate again

[tex]
(1 + x)e^x = \sum_{k=0}^{\infty} \frac{k^2 x^{k-1}}{k!}
[/tex]

Then, plug in [itex]x = 1[/itex] to get the sum.
 

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