Find the sum of the power series $\sum\limits_{n=1}^\infty \frac{(n+2)!}{(2!)(n!)}x^n$

frist I use $$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$$

and multiple two side by $x^2$

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

then diff each side two times

we can obtain $$\sum_{n=1}^\infty (n+1)(n+2)x^n= \frac{2x^3-6x^2+6x}{(1-x)^3}$$

but not the solution $=\frac{1}{(1-x)^3}$

  • $\begingroup$ use Newton's binomial theorem $\endgroup$ – user379195 May 11 '17 at 16:58
  • $\begingroup$ the result should be $$-\frac{x \left(x^2-3 x+3\right)}{(x-1)^3}$$ $\endgroup$ – Dr. Sonnhard Graubner May 11 '17 at 17:03
  • $\begingroup$ You should be using $\sum_{n=0}^{\infty}x^{n+2}=\frac{x^2}{(1-x)}$ then differentiate and which will get you $\frac{2x-x^2}{(1-x)^2}=n+2 \sum_{n=0}^{\infty}x^{n+1}$ then differentiate again and you'll get your final answer by finally diving by 2. $\endgroup$ – Iti Shree May 11 '17 at 17:06
  • $\begingroup$ thanks,I found that this question is wrong, n is start from zero. $\endgroup$ – Andrew Hung May 11 '17 at 17:27

Starting from zero,

$\begin{array}\\ \sum\limits_{n=0}^\infty \frac{(n+2)!}{(2!)(n!)}x^n &=\sum\limits_{n=0}^\infty \binom{n+2}{2}x^n\\ &=\frac1{x^2}(1-x)^{-3}\\ \end{array} $

by the generalized binomial theorem (since $\binom{-n}{k} =(-1)^n \binom{n+k-1}{k} $).


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