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

I'm trying to find the closed form for the specified series. However, I'm having a bit of trouble doing so. I assume there's a technique here that I haven't quite figured out, and any help would be appreciated.

Here's what I know:
$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$

And a technique on taking derivatives to find some other closed froms.

I'm not sure if I can apply any of these on the desired series.

HoopsMcCann
  • 895

6 Answers6

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

$\displaystyle = \sum_{n=0}^{\infty}[(2n+2)-1]x^{n}$

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

$\displaystyle = 2\sum_{n=0}^{\infty} \frac{\mathrm{d}}{\mathrm{d}x}x^{n+1} - \frac{1}{1-x}$

$\displaystyle = 2 \frac{\mathrm{d}}{\mathrm{d}x}\sum_{n=0}^{\infty} x^{n+1} - \frac{1}{1-x}$

$\displaystyle = 2 \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{x}{1-x} \right) - \frac{1}{1-x}$

$\displaystyle = \frac{2}{(1-x)^2} - \frac{1}{1-x}$

$\displaystyle = \frac{1+x}{(1-x)^2}$

Note that $|x| < 1$

PTDS
  • 3,464
2

Without calculus.

Start with $f_0(x) =\sum_{n=0}^{\infty} x^n =\dfrac1{1-x} $.

If $f_1(x) =\sum_{n=0}^{\infty} nx^n $, then $f_1(x)-xf_1(x) =(1-x)f_1(x) $ and

$\begin{array}\\ f_1(x)-xf_1(x) &=\sum_{n=0}^{\infty} nx^n-x\sum_{n=0}^{\infty} nx^n\\ &=\sum_{n=0}^{\infty} nx^n-\sum_{n=0}^{\infty} nx^{n+1}\\ &=\sum_{n=0}^{\infty} nx^n-\sum_{n=1}^{\infty} (n-1)x^{n}\\ &=\sum_{n=0}^{\infty} nx^n-\left(\sum_{n=1}^{\infty} nx^{n}-\sum_{n=1}^{\infty} x^{n}\right)\\ &=\sum_{n=1}^{\infty} x^{n}\\ &=f_0(x)-1\\ &=\dfrac1{1-x}-1\\ &=\dfrac{1-(1-x)}{1-x}\\ &=\dfrac{x}{1-x}\\ \end{array} $

so

$f_1(x) =\dfrac{x}{(1-x)^2} $.

Then

$\begin{array}\\ \sum_{n=0}^{\infty}(2n+1)x^n &=\sum_{n=0}^{\infty}(2n)x^n+\sum_{n=0}^{\infty}x^n\\ &=2\sum_{n=0}^{\infty}nx^n+f_0(x)\\ &=2\dfrac{x}{(1-x)^2}+\dfrac1{1-x}\\ &=\dfrac{2x+(1-x)}{(1-x)^2}\\ &=\dfrac{1+x}{(1-x)^2}\\ \end{array} $

marty cohen
  • 107,799
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HINT

To begin with, notice that

\begin{align*} \sum_{n=0}^{\infty}(2n+1)x^{n} = \sum_{n=0}^{\infty}[(2n+2)-1]x^{n} = 2\sum_{n=0}^{\infty}(n+1)x^{n} - \sum_{n=0}^{\infty}x^{n} \end{align*}

Then you make use of the fact that \begin{align*} \begin{cases} \displaystyle\sum_{n=0}^{\infty}x^{n} = \frac{1}{1-x} & \text{for}\,\,|x| < 1\\\\ \displaystyle (n+1)x^{n} = \frac{\mathrm{d}}{\mathrm{d}x}x^{n+1} \end{cases} \end{align*}

user0102
  • 21,572
1

see Help with summation: $\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$

Reusing that post result: $s_0 = \frac{x}{1-x}, s_1 = \frac{x}{(1-x)^2}$

$$\sum_{n=0}^\infty (2n+1) x^n = 1 + \sum_{n=1}^\infty (2n+1) x^n = 1 + 2 s_1 + s_0 = \frac{1+x}{(1-x)^2}$$

albert chan
  • 2,114
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Hint

$$\sum_{n=0}^{\infty} (2n+1)x^n=2x\sum_{n=0}^{\infty} nx^{n-1}+\sum_{n=0}^{\infty} x^{n}=2x \left(\sum_{n=0}^{\infty} x^{n} \right)'+\left(\sum_{n=0}^{\infty} x^{n} \right)$$

Claude Leibovici
  • 260,315
0

So basically we need to find , $$2 \sum_{n=0}^{\infty}nx^n + \sum_{n=0}^{\infty}x^n $$

First we'll solve for $ \sum_{n=0}^{\infty}x^n$ . Since $|x|<1$ $$ \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$

Now $\sum_{n=0}^{\infty}nx^n $ will be : $$ \sum_{n=0}^{\infty}nx^n=1x+2x^2+3x^3+.......\infty$$

So we need to solve for $1x+2x^2+3x^3+.......\infty$ this series . Let this series be defined as $\Lambda$

$$\Lambda=1x+2x^2+3x^3+.......\infty$$ $$\frac{\Lambda}{x}=1+2x+3x^2+.......\infty$$

Now , $$\frac{\Lambda}{x}-\Lambda=1+x+x^2+x^3+.....\infty$$ $$\Lambda\left(\frac{1-x}{x}\right)=\frac{1}{1-x}$$ $$\Lambda=\frac{x}{(1-x)^2}$$

FINALLY : $$2 \sum_{n=0}^{\infty}nx^n + \sum_{n=0}^{\infty}x^n =\frac{2x}{(1-x)^2}+\frac{1}{1-x}=\frac{1}{1-x}\left(\frac{2x+1-x}{1-x}\right)=\frac{x+1}{(x-1)^2}$$

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