2
$\begingroup$

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
    Why the downvotes?2012-03-21

3 Answers 3

1

I assume that the sum converges absolutely: $\sum_{n=0}^\infty(n+2)x^n=$ $=2(1+x+x^2+\ldots)+(x+x^2+x^3+\ldots)+(x^2+x^3+x^4+\ldots)+\ldots=$ $=(1+x+x^2+\ldots)(2+x+x^2+\ldots)=\frac{1}{1-x}\left(1+\frac{1}{1-x}\right)=$ $=\frac{2-x}{(1-x)^2}$

  • 0
    Nevermind, I see what you did there.2012-04-14
7

Hint:

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

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

  • 0
    @Trismegistos excellent remark.2012-03-21