3

According to Mathematica,

$$\sum_{k=0}^\infty \frac{(G+k)_{G-1}}{2^k}=2(G-1)!(2^{G}-1)$$

where

$$(G+k)_{G-1}=\frac{(G+k)!}{(G+k-G+1)!}=\frac{(G+k)!}{(k+1)!}$$

is the falling factorial. I would like to compute this analytically, but I have nothing I've been doing works. A proof by induction led me to a more complex summation, and I can split it or simplify the falling factorial. Is there any possible way to evaluate this without resorting to Mathematica? Any help and/or references would be greatly appreciated.

Marco Cantarini
  • 33,458
Joshuah Heath
  • 113

3 Answers3

2

Generating functions to the rescue! \begin{align*} \sum_{k=0}^\infty x^k &= \frac1{1-x} \\ \sum_{k=0}^\infty x^{G+k} &= \frac{x^G}{1-x} = \frac1{1-x} - 1 - x - \cdots - x^{G-1} \\ \frac{d^{G-1}}{dx^{G-1}} \sum_{k=0}^\infty x^{G+k} &= \frac{d^{G-1}}{dx^{G-1}} \bigg( \frac1{1-x} - 1 - x - \cdots - x^{G-1} \bigg) \\ \sum_{k=0}^\infty (G+k)_{G-1} x^{k+1} &= \frac{(G-1)!}{(1-x)^G} - (G-1)! \\ \sum_{k=0}^\infty (G+k)_{G-1} \big(\tfrac12\big)^{k+1} &= \frac{(G-1)!}{(1-\frac12)^G} - (G-1)! \\ \frac12 \sum_{k=0}^\infty \frac{(G+k)_{G-1}}{2^k} &= 2^G(G-1)! - (G-1)!. \end{align*} (The original series, and thus all subsequent series, have radius of convergence $1$, so plugging in $x=\frac12$ is valid.)

Greg Martin
  • 92,241
1

We have $$S=\sum_{k\geq0}\frac{\left(G+k\right)!}{\left(k+1\right)!}\left(\frac{1}{2}\right)^{k}=\left(G-1\right)!\sum_{k\geq0}\dbinom{G+k}{G-1}\left(\frac{1}{2}\right)^{k} $$ and since holds $$\dbinom{G+k}{G-1}=\sum_{m=0}^{G-1}\dbinom{k+m}{m} $$ we have, exchanging the sum with the series $$S=\left(G-1\right)!\sum_{m=0}^{G-1}\sum_{k\geq0}\dbinom{k+m}{m}\left(\frac{1}{2}\right)^{k} $$ now note that $$\frac{\left(k+m\right)!}{m!}=\left(k+m\right)\left(k+m-1\right)\cdots\left(k+m-\left(k-1\right)\right)=\left(-1\right)^{k}\left(-\left(m+1\right)\right)_{k}$$ where $\left(x\right)_{k}$ is the Pochhammer' symbol, so by the generalized binomial theorem we have $$\sum_{k\geq0}\dbinom{k+m}{m}\left(\frac{1}{2}\right)^{k}=\sum_{k\geq0}\dbinom{-\left(m+1\right)}{k}\left(-\frac{1}{2}\right)^{k}$$ $$=\frac{1}{\left(1-\frac{1}{2}\right)^{m+1}}=2^{m+1} $$ and finally $$\sum_{m=0}^{G-1}2^{m+1}=2\left(2^{G}-1\right)$$ so

$$\sum_{k\geq0}\frac{\left(G+k\right)!}{\left(k+1\right)!}\left(\frac{1}{2}\right)^{k}=2\left(G-1\right)!\left(2^{G}-1\right).$$

Community
  • 1
Marco Cantarini
  • 33,458
1

Assuming $G>0$, by using the Taylor series of the exponential function and the identity $m!=\int_{0}^{+\infty}x^m e^{-x}\,dx$ we have:

$$\begin{eqnarray*}\sum_{k\geq 0}\frac{(G+k)!}{2^k(k+1)!}&=&\sum_{k\geq 0}\frac{1}{2^k(k+1)!}\int_{0}^{+\infty}x^{G+k}e^{-x}\,dx\\[0.2cm]&=&\int_{0}^{+\infty}x^G e^{-x}\sum_{k\geq 0}\frac{x^k}{2^k(k+1)!}\,dx\\[0.2cm]&=&\int_{0}^{+\infty}2x^{G-1}(e^{-x/2}-e^{-x})\,dx\\[0.2cm]&=&2^{G+1}(G-1)!-2(G-1)!\\[0.2cm]&=&\color{red}{2(2^G-1)(G-1)!}\end{eqnarray*}$$

as wanted.

Jack D'Aurizio
  • 361,689