- #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

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- 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 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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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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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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[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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[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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[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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[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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[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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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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[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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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.

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]

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.

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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. :)

- #22

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