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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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marked as duplicate by Jack, tomasz, Jennifer Dylan, William, Nate Eldredge Aug 24 '12 at 4:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You could check out math.stackexchange.com/questions/120091/… and see if it helps. The same technique will apply to your term in $ix^{-i}$ Also math.stackexchange.com/questions/123023/… –  Ross Millikan Mar 21 '12 at 22:04
    
And see here. –  Bill Dubuque Mar 21 '12 at 22:19
    
So why we integrate ? & how you guys search by equation & find similar problems here ? –  xsari3x Mar 21 '12 at 22:27
    
Sorry , I mean differentiate –  xsari3x Mar 21 '12 at 22:30
    
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. –  Ross Millikan Mar 21 '12 at 23:09

3 Answers 3

up vote 1 down vote accepted

Let $y=x^{-1}$ (Thanks, Mark):

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

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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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Is there a name for a theory or something ? I really didn't get it :) –  xsari3x Mar 21 '12 at 22:19
    
@xsari3x did you notice that $d/dy(1+y+y^2+y^3+\dots)=1+2y+3y^2+\dots$ ? –  M. Alaggan Apr 1 '12 at 11:52
    
@M.Alaggan Yes , but ....Aha I deffe the Summation of the first one , to get the 2nd one –  xsari3x Apr 1 '12 at 19:53

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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It might be clearer if you added steps between the last two lines: $= 1/(1-1/x) + x^{-1}/(1-1/x) + x^{-2}/(1-1/x)$... $=(1+1/x+1/x^2+...)/(1-1/x)$. –  marty cohen Mar 22 '12 at 3:55
    
No harm meant in saying this, but "simple" is purely opinion. There can be elegance in simplicity, but mathematics, like life, is not always elegant and rarely simple. –  000 Mar 22 '12 at 20:53
    
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. –  Vadim Mar 23 '12 at 14:47

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