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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 :(

  • 1
    possible duplicate of [How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$](http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-1-infty-frac2n3n1)2012-03-21
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    not exactly the same. but Ill read that topic maybe it will be helpful2012-03-21
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    Why the downvotes?2012-03-21

3 Answers 3

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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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    I am not quite sure what you mean... sorry.2012-04-14
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    Nevermind, I see what you did there.2012-04-14
7

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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    oh...how come I didnt though of that? :) thank you2012-03-21
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    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.2012-03-21
5

, 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 you2012-03-21
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    @AmeliaYzaguirre thanks for the correction.2012-03-21
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    1. May be false. In general you can not rearrange infinite sums -- law is called [Riemann series theorem](http://mathworld.wolfram.com/RiemannSeriesTheorem.html)2012-03-21
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    @Trismegistos excellent remark.2012-03-21