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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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In summary: If you think you will be having trouble at small n, just write out the first few terms explicitly; for example, S =\sum_{n=0}^\infty x^n = 1 + x + x^2 + \sum_{n=3}^\infty x^n. Now it is perfectly clear what the first few terms of dS/dx \mbox{ and } d^2S/dx^2 ... are.

- #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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- #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]?

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Beer-monster said:

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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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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Beer-monster said:

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.

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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]

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Beer-monster said:

[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.

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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?

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Beer-monster said:

[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.

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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?

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Beer-monster said: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.

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Beer-monster said:

[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?

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Beer-monster said:

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 realize 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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Beer-monster said:

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

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Beer-monster said:

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 :)

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I agree. :)

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An infinite sum, also known as an infinite series, is a mathematical expression consisting of an infinite number of terms that are added together. It is often used to represent the sum of an infinite sequence of numbers.

The process of calculating an infinite sum involves finding the limit of the partial sums of the series. This is done by adding up a finite number of terms and then taking the limit as the number of terms approaches infinity. In some cases, the series may converge to a finite value, while in others it may diverge and not have a definite value.

A convergent infinite sum is one where the partial sums approach a finite value as the number of terms increases. In contrast, a divergent infinite sum is one where the partial sums do not approach a finite value, but either increase or decrease without bound.

Infinite sums have many applications in various fields of science, such as physics, engineering, and economics. For example, they can be used to calculate the total distance traveled by an object moving at a constant velocity, or to determine the total cost of a loan with a fixed interest rate.

Yes, there are several techniques that can be used to determine the convergence or divergence of an infinite sum, such as the comparison test, the ratio test, and the integral test. These tests involve comparing the given series to other known series with known convergence properties.

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