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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

Community
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xsari3x
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    You could check out http://math.stackexchange.com/questions/120091/how-do-we-get-the-result-of-the-summation-sum-limits-k-1n-k-cdot-2k/120100#120100 and see if it helps. The same technique will apply to your term in $ix^{-i}$ Also http://math.stackexchange.com/questions/123023/how-to-calculate-sum-n-0-infty-n2xn – 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 Answers3

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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.

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

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

000
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The simplest way is as follows: $$\sum_{i=0}^\infty(i+1)x^{-i}=1+2x^{-1}+3x^{-2}+\ldots$$ $$=(1+x^{-1}+x^{-2}+\ldots)+(x^{-1}+x^{-2}+\ldots)+(x^{-2}+\ldots)$$ $$=(1+x^{-1}+x^{-2}+\ldots)(1+x^{-1}+x^{-2}+\ldots)=\frac{1}{(1-x^{-1})^2}$$

Vadim
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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 03:55
  • 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