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Possible Duplicate:
How to calculate $\sum_{n=0}^\infty {(n+2)}x^{n}$

I'm sorry if I'm asking in wrong title .. I'm not a math expert ...

I need to know the rule behind this problem & how did it ended like this :)

$\sum_{i=0}^\infty (i+1)x^{-i} = \frac{1}{(1-x^{-1})^2}$

I'm ready to clarify any question you need to ask .

Thanks in advance

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    I found the other questions because I had answered one in the past and seen the other recently. If you click on your name at the top of the page you can see all your activity.2012-03-21

3 Answers 3

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The simplest way is as follows: $\sum_{i=0}^\infty(i+1)x^{-i}=1+2x^{-1}+3x^{-2}+\ldots$ $=(1+x^{-1}+x^{-2}+\ldots)+(x^{-1}+x^{-2}+\ldots)+(x^{-2}+\ldots)$ $=(1+x^{-1}+x^{-2}+\ldots)(1+x^{-1}+x^{-2}+\ldots)=\frac{1}{(1-x^{-1})^2}$

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    By simplest I meant two things: a) This method does not require any additional advanced methods such as differentiating series (see * below); b) It is one of the best method to quickly get the answer for such series even without pen and paper(compare to any other you have seen). * That being said, I must add that it also requires one to check whether you can rearrange the terms in the sum. Likely, in our case this thing converges iff it converges absolutely: for |x|>1. I do not think any other solution, including differentiating, even mentions the conditions under which it works.2012-03-23
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If you put $y=x^{-1}$ then you are summing $1+2y+3y^2+...$

Then you could try summing $1+y+y^2+y^3+...$ and differentiating both sides.

You will have to do a little work to verify convergence, if that is a significant consideration for you.

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    @M.Alaggan Yes , but ....Aha I deffe the Summation of the first one , to get the 2nd one2012-04-01
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Let $y=x^{-1}$ (Thanks, Mark):

$S=\sum_{i=0}^{\infty}(i+1)y^i=\sum_{i=0}^{\infty}iy^i+\sum_{i=0}^{\infty}y^i$ $u=\sum_{i=0}^{\infty}iy^i=0+y+2y^2+\dots$ $u-yu=0+y+2y^2+\dots-(0+y^2+2y^3+\dots)=y+y^2+\dots $ $y+y^2+\dots=z$ $z-yz=y+y^2+\dots-(y^2+y^3+\dots)=y$ $z=\frac{y}{1-y}$ $u-yu=\frac{y}{1-y} \Rightarrow u=\frac{y}{(1-y)^2}$ $\sum_{i=0}^{\infty}y^i=1+z$ $\therefore S=1+\frac{y}{1-y}+\frac{y}{(1-y)^2}=\frac{(1-y)^2+y(1-y)+y}{(1-y)^2}=\frac{(1-y)^2+y(2-y)}{(1-y)^2}=\frac{1}{(1-y)^2}$