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

Question

I need help to find the closed form of the following power series: $$\sum_{n=1}^\infty (n^2+n)(x-1)^n$$

Edit: I have added a solution in the answers below :)

Related: https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-inftyn1xn — Arctic Char, Dec 09 '19 at 04:58
if you dont show some work, some nice users will vote to close your post. try to add some work . — Ali Shadhar, Dec 09 '19 at 05:52

gt6989b · Answer 1 · 2019-12-09T05:31:35.640

1

HINT Note that you have $$ \sum_{n=1}^\infty (n^2+n)(x-1)^n = (x-1)^1 \sum_{n=1}^\infty n(n+1)(x-1)^{n-1} $$ and the sum can be recognized as a second derivative of something very familiar...

edited Dec 09 '19 at 05:31

answered Dec 09 '19 at 05:18

gt6989b

54,422

@StrugglingStudent made a mistake, please see the corrected version. – gt6989b Dec 09 '19 at 05:32

score 0 · Answer 2 · answered Dec 09 '19 at 05:37

Let $X=x-1$ and consider $$S=\sum_{n=1}^\infty (n^2+n)X^n$$ The trick being to write $$n^2=n(n-1)+n$$ you then face $$S=\sum_{n=1}^\infty n(n-1) X^n+2\sum_{n=1}^\infty n X^n=X^2\sum_{n=1}^\infty n(n-1) X^{n-2}+2X\sum_{n=1}^\infty n X^{n-1}$$ that is to say $$S=X^2 \left(\sum_{n=1}^\infty X^{n}\right)''+2X\left(\sum_{n=1}^\infty X^{n}\right)'$$

score 0 · Answer 3 · answered Dec 09 '19 at 05:51

Based on other answers, I think I have gotten the correct solution and verified it with Mathematica.

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

And using the Taylor series of $\frac{1}{1-x}$ about $x_0 = 1$ and its first and second derivatives, you get $$f(x) = (x-1)^2*\frac{2}{(2-x)^2} + \frac{2(x-1)}{(2-x)^2} = \frac{2x-2}{(2-x)^3}$$

score 0 · Answer 4 · answered Dec 09 '19 at 06:03

Start with

$$\sum_{n=1}^\infty z^n=\frac{z}{1-z},\quad |z|<1$$

differentiate both sides

$$\sum_{n=1}^\infty nz^{n-1}=\frac{1}{(1-z)^2}$$

now multiply both sides by $z^2$ then differentiate

$$\sum_{n=1}^\infty (n^2+n)z^n=\frac{2z}{(1-z)^3}$$

set $z=x-1$

$$\sum_{n=1}^\infty (n^2+n)(x-1)^n=\frac{2(x-1)}{(2-x)^3}$$

score 0 · Answer 5 · answered Dec 09 '19 at 06:25

$$(1-z)^{-1}=\sum_{k=0}^{infty} z^k ,~ |z|<1~~~(1)$$ Differentiaate w.r.t.xz, we get $$\frac{z}{(1-z)^2}= \sum k z^{k}~~~~(2)$$ Multiply (2) by $z^2$ and differentiate w. r. t. $z$, we get $$\sum_{k=0}^{\infty} k(k+1) z^k =\frac{2z}{(1-z)^3}~~~~(3)$$ Now, by putting $z=(x-1)$ in (3), we get $$\sum_{k=0}^{\infty} k(k+1)(x-1)^k= \frac{2(x-1)}{(2-x)^3},~~ 0<x <2.$$

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

5 Answers5