- #1

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[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

- Thread starter Beer-monster
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- #1

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[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

- #2

Dick

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- #3

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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 did you get the sum of n*x^(n-1)? What is it?

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]?

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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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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.

i.e. [tex] \frac{d1}{dx}+\frac{dx}{dx}+\frac{d^{2}x^{2}}{dx^{2}} ...\frac{d^{n}x^{n}}{dx^{n}} [/tex]

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[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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Ok, so sum n=0 to infinity n*x^(n-1)=1/(1-x)^2. Right? Now take a second derivative.

[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]

- #9

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[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'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.

[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?

- #11

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[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

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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.

- #14

Dick

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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?

[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?

- #15

Ray Vickson

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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.

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.

RGV

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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]

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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Closer and closer. Now you've got sum n*x^n ok. But you don't really mean x^3, do you? Keep checking.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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- #19

Dick

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Maybe, so what's your final answer? I'm dying to know.

- #20

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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. :)

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