Proving that $\sum_{n=1}^{\infty}S(n,n-2)x^n = \frac{x^3(1+2x)}{(1-x)^5}$

Question

I was wondering if anyone could give a hint on how to prove this expression, I have been stuck on it for hours. Thanks in advance!

Proving that $$\sum_{n=1}^{\infty}S(n,n-2)x^n = \frac{x^3(1+2x)}{(1-x)^5}$$ where $S(n,n-2)$ are the Stirling numbers of the second kind.

Answer 1

Given the known recurrence for Stirling No. of the $2$nd kind $$ \left\{ \matrix{ n \cr m \cr} \right\} = m\left\{ \matrix{ n - 1 \cr m \cr} \right\} + \left\{ \matrix{ n - 1 \cr m - 1 \cr} \right\} $$ we obtain $$ \eqalign{ & \left\{ \matrix{ n \cr n - 1 \cr} \right\} = \left( {n - 1} \right)\left\{ \matrix{ n - 1 \cr n - 1 \cr} \right\} + \left\{ \matrix{ n - 1 \cr n - 2 \cr} \right\} = \left( {n - 1} \right) + \left\{ \matrix{ n - 1 \cr n - 2 \cr} \right\}\quad \Rightarrow \quad \cr & \Rightarrow \quad \left\{ \matrix{ n \cr n - 1 \cr} \right\} = \left( \matrix{ n \cr n - 2 \cr} \right) = \left[ {0 \le n} \right]\left( \matrix{ n \cr 2 \cr} \right) \cr} $$ where $[P]$ is the Iverson bracket: $ \left\{ \matrix{ [TRUE] = 1 \hfill \cr [FALSE] = 0 \hfill \cr} \right. $

and then $$ \eqalign{ & \left\{ \matrix{ n \cr n - 2 \cr} \right\} = \left( {n - 2} \right)\left\{ \matrix{ n - 1 \cr n - 2 \cr} \right\} + \left\{ \matrix{ n - 1 \cr n - 3 \cr} \right\} = \cr & = \left( {n - 2} \right)\left( \matrix{ n - 1 \cr n - 3 \cr} \right) + \left\{ \matrix{ n - 1 \cr n - 3 \cr} \right\} = \left[ {1 \le n} \right]\left( {n - 2} \right)\left( \matrix{ n - 1 \cr 2 \cr} \right) + \left\{ \matrix{ n - 1 \cr n - 3 \cr} \right\} \cr} $$

Therefrom we can write (we can take the summation index to start from $0$) $$ \eqalign{ & F(x) = \sum\limits_{0\, \le \,n} {\left\{ \matrix{ n \cr n - 2 \cr} \right\}x^{\,n} } = \cr & = \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{ n - 1 \cr n - 3 \cr} \right)x^{\,n} } + x\sum\limits_{0\, \le \,n} {\left\{ \matrix{ n - 1 \cr n - 3 \cr} \right\}x^{\,n - 1} } = \cr & = \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{ n - 1 \cr n - 3 \cr} \right)x^{\,n} } + x\,F(x) \cr} $$

which finally gives $$ \eqalign{ & \left( {1 - x} \right)F(x) = \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{ n - 1 \cr n - 3 \cr} \right)x^{\,n} } = x^{\,3} \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{ n - 1 \cr n - 3 \cr} \right)x^{\,n - 3} } = \cr & = x^{\,3} {d \over {d\,x}}\sum\limits_{0\, \le \,n} {\left( \matrix{ n - 1 \cr n - 3 \cr} \right)x^{\,n - 2} } = x^{\,3} {d \over {d\,x}}{1 \over x}\sum\limits_{1\, \le \,n} {\left( \matrix{ n - 1 \cr 2 \cr} \right)x^{\,n - 1} } = x^{\,3} {d \over {d\,x}}{1 \over x}{{x^{\,2} } \over {\left( {1 - x} \right)^{\,3} }} = \cr & = x^{\,3} {d \over {d\,x}}{x \over {\left( {1 - x} \right)^{\,3} }} = x^{\,3} {{1 + 2x} \over {\left( {1 - x} \right)^{\,4} }} \cr} $$

Q.E.D.

Answer 2

How are you defining the Stirling numbers of the second kind? What identities or recurrences involving them do you know? It happens that $$ S(n,n-2) = \dfrac{(3n-5) n (n-1)(n-2)}{24} $$ Or perhaps something else at OEIS sequence A001296 may be useful.