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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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possible duplicate of How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$ – Did Mar 21 '12 at 20:05
not exactly the same. but Ill read that topic maybe it will be helpful – YNWA Mar 21 '12 at 20:08
Why the downvotes? – user7530 Mar 21 '12 at 20:32
up vote 4 down vote accepted

, 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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didnt even crossed my mind to try two different series. thank you – YNWA Mar 21 '12 at 20:36
@AmeliaYzaguirre thanks for the correction. – user2468 Mar 21 '12 at 22:27
1. May be false. In general you can not rearrange infinite sums -- law is called Riemann series theorem – Trismegistos Mar 21 '12 at 22:49
@Trismegistos excellent remark. – user2468 Mar 21 '12 at 23:20


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

share|cite|improve this answer come I didnt though of that? :) thank you – YNWA Mar 21 '12 at 20:36
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. – GEdgar Mar 21 '12 at 21:09

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

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