Just need a quick explanation (Power Series)

[DivGradCurl]

Problem:

(a) Expand

$$f(x)=\frac{x+x^2}{\left( 1-x\right) ^3}$$

as a power series.

(b) Use part (a) to find the sum of the series

$$\sum _{n=1} ^{\infty} \frac{n^2}{2^n}$$

$$\hline$$

Solution:

(a)

$$f(x)=\frac{x+x^2}{\left( 1-x\right) ^3} = \left( x + x^2 \right) \left[ 1 + (-x) \right] ^{-3} = \left( x + x^2 \right) \sum _{n=0} ^{\infty} \binom{-3}{n} (-x) ^n = \left( x + x^2 \right) \sum _{n=0} ^{\infty} (-1) ^n \binom{-3}{n} x ^n$$

$$f(x) = \left( x + x^2 \right) \left[ \binom{-3}{0} - \binom{-3}{1}x + \binom{-3}{2} x^2 - \binom{-3}{3} x^3 + \binom{-3}{4} x^4 - \cdots \right]$$

$$f(x) = \left( x + x^2 \right) \left[ 1 - \frac{(-3)}{1!}x + \frac{(-3)(-4)}{2!}x^2 - \frac{(-3)(-4)(-5)}{3!}x^3 + \frac{(-3)(-4)(-5)(-6)}{4!}x^4 -\cdots \right]$$

$$f(x) = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} \frac{\left( n+2 \right)!}{n!}x^n \right] = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} (n+2)(n+1)x^n \right]$$

$$f(x) = x + x^2 + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+1} \right] + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+2} \right]$$

$$f(x) = x + x^2 - \frac{x\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} - \frac{x^2\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} \qquad \fbox{UNCLEAR}$$

$$f(x) = x + x^2 - \left[ x +\frac{1}{\left( x-1 \right) ^2} +\frac{1}{\left( x-1 \right) ^3} \right] - \left[ x^2 + \frac{1}{ x-1} + \frac{2}{\left( x-1 \right) ^2} + \frac{1}{\left( x-1 \right) ^3} \right]$$

$$f(x) = -\frac{1}{ x-1} -\frac{3}{\left( x-1 \right) ^2} - \frac{2}{\left( x-1 \right) ^3} = \frac{x\left( x+1 \right)}{\left( x-1 \right) ^3} = x \frac{d}{dx} x \frac{d}{dx} \left[ \frac{1}{1-x} \right] = x \frac{d}{dx} x \frac{d}{dx} \left[ \sum _{n=0} ^{\infty} x^n \right]$$

$$f(x) = \sum _{n=1} ^{\infty} n^2 x^n$$

(b)

$$\sum _{n=1} ^{\infty} \frac{n^2}{2^n} = \sum _{n=1} ^{\infty} n^2 \left( \frac{1}{2} \right) ^n = f\left( \frac{1}{2} \right) = \frac{\frac{1}{2}+\left( \frac{1}{2} \right)^2}{\left( 1-\frac{1}{2}\right) ^3} = 6$$

Comments:

$$\fbox{UNCLEAR}:$$ I don't understand the transition from the series above to this result.

Thank you

 

[AKG]

You know

$$g(x) = \sum _{n = 0} ^{\infty}\frac{g^{(n)}(a)}{n!}(x-a)^n$$

Can you find some g such that you can express:

$$\sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+1} \right]$$

in the form above?

 

[DivGradCurl]

Thanks for the tip!

(a)

$$f(x)=\frac{x+x^2}{\left( 1-x\right) ^3} = \left( x + x^2 \right) \left[ 1 + (-x) \right] ^{-3} = \left( x + x^2 \right) \sum _{n=0} ^{\infty} \binom{-3}{n} (-x) ^n = \left( x + x^2 \right) \sum _{n=0} ^{\infty} (-1) ^n \binom{-3}{n} x ^n$$

$$f(x) = \left( x + x^2 \right) \left[ \binom{-3}{0} - \binom{-3}{1}x + \binom{-3}{2} x^2 - \binom{-3}{3} x^3 + \binom{-3}{4} x^4 - \cdots \right]$$

$$f(x) = \left( x + x^2 \right) \left[ 1 - \frac{(-3)}{1!}x + \frac{(-3)(-4)}{2!}x^2 - \frac{(-3)(-4)(-5)}{3!}x^3 + \frac{(-3)(-4)(-5)(-6)}{4!}x^4 -\cdots \right]$$

$$f(x) = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} \frac{\left( n+2 \right)!}{n!}x^n \right] = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} (n+2)(n+1)x^n \right]$$

$$f(x) = x + x^2 + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+1} \right] + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+2} \right]$$

$$f(x) = x + x^2 + \left\{ \frac{x}{2} \frac{d^2}{dx^2} \left[ x^2 \sum _{n=0} ^{\infty} x^n \right] - x \right\} + \left\{ \frac{x^2}{2} \frac{d^2}{dx^2} \left[ x^2 \sum _{n=0} ^{\infty} x^n \right] - x^2 \right\}$$

$$f(x) = x + x^2 + \left\{ \frac{x}{2} \frac{d^2}{dx^2} \left[ x^2 \left( \frac{1}{1-x} \right) \right] - x \right\} + \left\{ \frac{x^2}{2} \frac{d^2}{dx^2} \left[ x^2 \left( \frac{1}{1-x} \right) \right] - x^2 \right\}$$

$$f(x) = x + x^2 - \frac{x\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} - \frac{x^2\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3}$$

$$f(x) = x + x^2 - \left[ x +\frac{1}{\left( x-1 \right) ^2} +\frac{1}{\left( x-1 \right) ^3} \right] - \left[ x^2 + \frac{1}{ x-1} + \frac{2}{\left( x-1 \right) ^2} + \frac{1}{\left( x-1 \right) ^3} \right]$$

$$f(x) = -\frac{1}{ x-1} -\frac{3}{\left( x-1 \right) ^2} - \frac{2}{\left( x-1 \right) ^3} = \frac{x\left( x+1 \right)}{\left( x-1 \right) ^3} = x \frac{d}{dx} x \frac{d}{dx} \left[ \frac{1}{1-x} \right] = x \frac{d}{dx} x \frac{d}{dx} \left[ \sum _{n=0} ^{\infty} x^n \right]$$

$$f(x) = \sum _{n=1} ^{\infty} n^2 x^n$$

(b)

$$\sum _{n=1} ^{\infty} \frac{n^2}{2^n} = \sum _{n=1} ^{\infty} n^2 \left( \frac{1}{2} \right) ^n = f\left( \frac{1}{2} \right) = \frac{\frac{1}{2}+\left( \frac{1}{2} \right)^2}{\left( 1-\frac{1}{2}\right) ^3} = 6$$