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$$\sum_0^\infty(n+1)(2n+1)x^n$$

I know the following, which is what I am assuming I must manipulate. I have the answer to the closed form, but I do not understand how to get there. Please, no answers as of yet, just hints/tips. $$ \sum_0^\infty(2n+1)x^{2n}=\frac{1+x^2}{(1-x^2)^2} $$

H5159
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  • Start with $\sum_{n \ge 0} x^n = \frac{1}{1-x}$. Differentiate the left and right. What do you get? (If the exponent of $x$ seems less by one from what you'd like, you can then multiply by $x$ on both sides.) And what if you differentiate again? – ShreevatsaR Mar 22 '14 at 03:50
  • You would get the following two: $$\sum_0^\infty nx^n=\frac{x}{(1-x)^2}$$ $$ \sum_0^\infty (n-1)(n)x^n = \frac{-(x+1)}{(1-x)^3} $$ – H5159 Mar 22 '14 at 03:51
  • $$ \sum_0^\infty (n-1)(n)x^n = \frac{-x(x+1)}{(1-x)^3} $$ Edited ^ – H5159 Mar 22 '14 at 03:57
  • Cool, you've got an expression for $\sum nx^n$; can you do something again and get an expression for $\sum n^2x^n$? Then note that $(n+1)(2n+1) = 2n^2 + 3n + 1$. – ShreevatsaR Mar 22 '14 at 04:02
  • You would take the derivative and multiply by x again to get $\sum_0^\infty n^2x^n$ $$\sum n^2x^n = \frac{-x(x+1)}{(x-1)^3}$$ Unless I am doing these wrong, I don't see where we can use $2n^2+3n+1$. – H5159 Mar 22 '14 at 04:04
  • $\sum (n+1)(2n+1) x^n = \sum (2n^2 + 3n + 1) x^n = \sum 2n^2 x^n + \sum 3n x^n + \sum x^n$. Does that help? – ShreevatsaR Mar 22 '14 at 04:22

3 Answers3

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Hint

Since you have $(n+1)(2n+1)x^n$ I suppose it is natural to consider that first and second derivatives, as well as function, would appear in the result.

Since $(n+1)(2n+1)=2n^2+3n+1=2n(n-1)+5n+1$, we can then rewrite $$\sum_0^\infty(n+1)(2n+1)x^n=2x^2 \sum_0^\infty n(n-1)x^{n-2}+ 5 x \sum_0^\infty n x^{n-1}+ \sum_0^\infty x^{n}$$ where appear $$\sum_0^\infty x^{n}=\frac{1}{1-x}$$ as well as its first and second derivative with respect to $x$.

I am sure that you can take from here and finish.

Claude Leibovici
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The general recipe is if $f(x) = \sum\limits_{n=0}^\infty a_n x^n$, then

$$ (x\frac{d}{dx})^kf(x) = \sum_{n=0}^{\infty} n^ka_n x^n \quad\text{ and }\quad \frac{1}{x^k}\int_0^x t^{k-1} f(t) dt = \sum_{n=0}^\infty \frac{a_n}{n+k}x^n $$ If you have a sequence whose coefficients are coefficients of something function $f(x)$ you know times some polynomial in $n$, then you can obtain the generating function by applying corresponding polynomial in $(x\frac{d}{dx})$ to $f(x)$. e.g.

$$\begin{align} \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \quad\implies\quad \sum_{n=0}^{\infty} (n+1)(2n+1)x^n = & (x\frac{d}{dx} + 1)(2x\frac{d}{dx}+1)\frac{1} {1-x}\\ = & (x\frac{d}{dx} + 1)\frac{1+x}{(1-x)^2}\\ = & \frac{1+3x}{(1-x)^3} \end{align} $$

achille hui
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Try the following strategy:

step 1: replace x^2 by y on both sides.

step 2: multiply both sides by y.

step 3: differentiate both sides with respect to y.

step 4: replace y by x again and finish.

DeepSea
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  • Doing this does match the closed form which I had. But I do not understand how you knew to do this. What was your inspiration? – H5159 Mar 22 '14 at 04:01
  • Actually, it looks like there is a negative being multiplied through. My closed form that I'm supposed to get seemed to be multiplied by -1 to get what you've suggested. – H5159 Mar 22 '14 at 04:13
  • @Frumpy: Start out by manipulating the geometric sum 1 + x + x^2 +.. + x^n = (1 - x^n+1)/(1 - x) and differentiate it and remultiply. I actually learned it here like you. – DeepSea Mar 22 '14 at 04:13