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I want to calculate the sum of $$\sum_{n=0}^\infty {(n+2)}x^{n}$$

I have tried to look for a known taylor/maclaurin series to maybe integrate or differentiate...but I did not find it :|

Thank you.

edit : i see a similarity to $\frac{1}{1-x}$ but I dont know how to go from there :(

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Hint:

$$\rm (n+2)x^n=\frac{d}{dx}\big(x^{n+1}\big)+x^n, \qquad \sum_{n=0}^\infty x^{n+k}=\frac{x^k}{1-x}$$

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  • $\begingroup$ oh...how come I didnt though of that? :) thank you $\endgroup$
    – YNWA
    Mar 21 '12 at 20:36
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    $\begingroup$ Alternate: Multiply by $x$ to get $(n+2)x^{n+1}$, then integrate to get $x^{n+2}$. That way you get one series that can be recognized. $\endgroup$
    – GEdgar
    Mar 21 '12 at 21:09
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, so few hints:

  1. $\sum_{n = 0}^{\infty}(n+2)x^n = \sum_{n = 0}^{\infty}nx^n + 2\sum_{n = 0}^{\infty}x^n $

  2. $\frac{1}{1-x} = 1 + x + x^2 + \ldots$

  3. $\frac{d}{dx} (\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}) \to (\sum_{n=0}^{\infty} \color{red}{??}x^{\color{red}{??}} = \color{red}{??})$

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  • $\begingroup$ didnt even crossed my mind to try two different series. thank you $\endgroup$
    – YNWA
    Mar 21 '12 at 20:36
  • $\begingroup$ @AmeliaYzaguirre thanks for the correction. $\endgroup$
    – user2468
    Mar 21 '12 at 22:27
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    $\begingroup$ 1. May be false. In general you can not rearrange infinite sums -- law is called Riemann series theorem $\endgroup$ Mar 21 '12 at 22:49
  • $\begingroup$ @Trismegistos excellent remark. $\endgroup$
    – user2468
    Mar 21 '12 at 23:20
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I assume that the sum converges absolutely: $$\sum_{n=0}^\infty(n+2)x^n=$$ $$=2(1+x+x^2+\ldots)+(x+x^2+x^3+\ldots)+(x^2+x^3+x^4+\ldots)+\ldots=$$ $$=(1+x+x^2+\ldots)(2+x+x^2+\ldots)=\frac{1}{1-x}\left(1+\frac{1}{1-x}\right)=$$ $$=\frac{2-x}{(1-x)^2}$$

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  • $\begingroup$ I am not quite sure what you mean... sorry. $\endgroup$
    – Vadim
    Apr 14 '12 at 17:11
  • $\begingroup$ Nevermind, I see what you did there. $\endgroup$
    – Pedro Tamaroff
    Apr 14 '12 at 20:21

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