Help with infinite sum.

  • #1
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Could someone suggest how I could go about calculating this infinite sum?

[tex]\sum_{n=1}^{\infty}n^{2}x^{n}[/tex]

I'm having some trouble as this series does not have a constant ratio, difference or matches any standard series I can think of at the moment.

Thanks
 

Answers and Replies

  • #2
Dick
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You derive something like that by looking at derivatives of the geometric series. You can sum x^n, n=0 to infinity, right? Take a derivative of both sides. What's sum n*x^n, k=0 to infinity? Now take another derivative.
 
  • #3
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So, take the derivative, do the sum then take the new derivative?

So the first sum would be the derivative of [itex] nx^{n-1} [/itex]

Or just look at the what derivatives I can solve that give me [itex]n^{2}x^{n}[/itex]?
 
  • #4
Dick
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So, take the derivative, do the sum then take the new derivative?

So the first sum would be the derivative of [itex] nx^{n-1} [/itex]

Or just look at the what derivatives I can solve that give me [itex]n^{2}x^{n}[/itex]?
So did you get the sum of n*x^(n-1)? What is it?
 
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  • #5
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It's the sum of increasing derivatives of [itex]x^{n}[/itex].

i.e. [tex] \frac{d1}{dx}+\frac{dx}{dx}+\frac{d^{2}x^{2}}{dx^{2}} ...\frac{d^{n}x^{n}}{dx^{n}} [/tex]
 
  • #6
Dick
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It's the sum of increasing derivatives of [itex]x^{n}[/itex].

i.e. [tex] \frac{d1}{dx}+\frac{dx}{dx}+\frac{d^{2}x^{2}}{dx^{2}} ...\frac{d^{n}x^{n}}{dx^{n}} [/tex]
No, that's not what I mean and it's not right. sum x^n=1/(1-x), right? Differentiate BOTH sides of the equation.
 
  • #7
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As so:

[tex] \frac{d}{dx}\sum{x^{n}}=\sum \frac{d}{dx}x^n = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^{2}} [/tex]
 
  • #8
Dick
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As so:

[tex] \frac{d}{dx}\sum{x^{n}}=\sum \frac{d}{dx}x^n = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^{2}} [/tex]
Ok, so sum n=0 to infinity n*x^(n-1)=1/(1-x)^2. Right? Now take a second derivative.
 
  • #9
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You mean

[tex] \sum n(n-1)x^{n-2} = \frac{-2}{(1-x)^{3}} [/tex]

Then do I remove the extra factor of x^{n-2} by adding on both sides to get my final answer?
 
  • #10
Dick
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You mean

[tex] \sum n(n-1)x^{n-2} = \frac{-2}{(1-x)^{3}} [/tex]

Then do I remove the extra factor of x^{n-2} by adding on both sides to get my final answer?
You've got a sign error in that derivative. If you know sum n(n-1)x^(n-2) then you can find sum n(n-1)x^n, right? They differ by a factor of x^2. Now n(n-1)*x^n=n^2*x^n-n*x^n. That's the general strategy. You'll need to pay some attention to the details of the limits to get the right answer.
 
  • #11
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I think I get the idea, but I'm getting something messy out.

[tex] \frac{2x^{2}}{(1-x)^{3}}+\frac{1}{(1-x)} [/tex]

To take care of the fact that the sum is between 1 and infinity not 0 and infinity can I just subtract the n=0 term from my calculated sum?
 
  • #12
Ray Vickson
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I think I get the idea, but I'm getting something messy out.

[tex] \frac{2x^{2}}{(1-x)^{3}}-\frac{1}{(1-x)} [/tex]

To take care of the fact that the sum is between 1 and infinity not 0 and infinity can I just subtract the n=0 term from my calculated sum?

Well, just look at the n=0 term in sum n*x^(n-1), etc. You really can answer your questions for yourself.

RGV
 
  • #13
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Yes. I realise that came out a bit more stupid than I intended it to. My apologies for not being specific enough.


However, I was mostly thinking in general terms. I have a few more sums to work out, and I hope to use this strategy in more situations. To subtract the n=0 term (should it be non-zero) from a the result of the standard infinite to get the sum between n=1 and n=infinite makes sense to me, but I've often been wrong about these things before.:smile:
 
  • #14
Dick
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I think I get the idea, but I'm getting something messy out.

[tex] \frac{2x^{2}}{(1-x)^{3}}+\frac{1}{(1-x)} [/tex]

To take care of the fact that the sum is between 1 and infinity not 0 and infinity can I just subtract the n=0 term from my calculated sum?
You are getting pretty close. sum n^2*x^n-sum n*x^n=2x^2/(1-x)^3, right? But I think you've got the sum n*x^n part wrong. Can you go back and fix that?
 
  • #15
Ray Vickson
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Yes. I realise that came out a bit more stupid than I intended it to. My apologies for not being specific enough.


However, I was mostly thinking in general terms. I have a few more sums to work out, and I hope to use this strategy in more situations. To subtract the n=0 term (should it be non-zero) from a the result of the standard infinite to get the sum between n=1 and n=infinite makes sense to me, but I've often been wrong about these things before.:smile:
If you think you will be having trouble at small n, just write out the first few terms explicitly; for example, [itex]S =\sum_{n=0}^\infty x^n = 1 + x + x^2 + \sum_{n=3}^\infty x^n. [/itex] Now it is perfectly clear what the first few terms of [itex] dS/dx \mbox{ and } d^2S/dx^2 [/itex] are.

RGV
 
  • #16
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Sorry, I realise now I wrote down my sum wrong on my paper.

Using
[tex] \sum nx^{n-1} = \frac{1}{(1-x)^{2}} [/tex]

[tex] \sum nx^{n}= \frac{x}{(1-x)^{2}}[/tex]

Thus my answer is:

[tex]\frac{2x^{2}}{(1-x)^{3}}+\frac{x^{3}}{(1-x)^{2}} [/tex]
 
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  • #17
Dick
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Sorry, I realise now I wrote down my sum wrong on my paper.

Using
[tex] \sum nx^{n-1} = \frac{1}{(1-x)^{2}} [/tex]

[tex] \sum nx^{n}= \frac{x}{(1-x)^{2}}[/tex]

Thus my answer is:

[tex]\frac{2x^{2}}{(1-x)^{3}}+\frac{x^{3}}{(1-x)^{2}} [/tex]
Closer and closer. Now you've got sum n*x^n ok. But you don't really mean x^3, do you? Keep checking.
 
  • #18
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No, the x^2s cancel. This is becoming a vivid reminder of how I shouldn't try and rush maths in my head and should sit down with a pencil and shut up :)
 
  • #19
Dick
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No, the x^2s cancel. This is becoming a vivid reminder of how I shouldn't try and rush maths in my head and should sit down with a pencil and shut up :)
Maybe, so what's your final answer? I'm dying to know.
 
  • #20
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[tex]\frac{2x^{2}}{(1-x)^{3}}+\frac{x}{(1-x)^{2}} [/tex]

Which can be simplified to:

[tex] \frac{x(1+x)}{(1-x)^{3}}[/tex]

Which makes sense in the problem I'm doing, and maple agrees :)
 
  • #21
Dick
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I agree. :)
 
  • #22
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Thanks for all your help. I had to use this strategy again for further sums in the problem I was working on so going over it was really helpful.
 

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