As part of giving a computationally efficient means of calculating the gamma distribution, Williams in 'Weighing the Odds', pg 149 asserts the following identity, where $\Gamma$ denotes the Gamma function $\left(~\mbox{recall that}\ \Gamma\left(K + 1\right) = K\,\Gamma\left(K\right)~\right)$:
\begin{align} \mbox{Let}\quad {\rm h}\left(x\right) &=\sum_{n=0}^{\infty}\frac{x^{n}}{\Gamma\left(K + n + 1\right)}. \\[2mm]\mbox{The claim is that}\quad x{\rm h}'\left(x\right)&= \frac{1}{\Gamma\left(K\right)} + \left(\,x - K\,\right){\rm h}\left(x\right). \end{align}
I'm a bit at a loss on how to approach this. Normally with power series I'd try to prove the left and right hand sides have the same coefficients, but that cannot possibly be the case here since there is a constant term on the right hand side, and not the left.
Any ideas ?
Using the identity that $\Gamma(K+1)=K\Gamma(K)$ repeatedly, for integer $m$ $$\Gamma(K+m)=(K+m-1)\Gamma(K+m-1)\cdots=\Gamma(K)\prod_{g=0}^{m-1}(K+g)$$ we explicitly express the first few terms of $(x-K)h(x)$ as follows:- $$(x-K)h(x)=\frac{(x-K)}{K\Gamma(K)}+\frac{x(x-K)}{(K+1)K\Gamma(K)}+\frac{x^2(x-K)}{(K+2)(K+1)K\Gamma(K)}+\cdots\\=-\frac{1}{\Gamma(K)}+\frac{x}{(K+1)K\Gamma(K)}+\frac{2x^2}{(K+2)(K+1)K\Gamma(K)}+\cdots+\frac{mx^m}{\Gamma(K)\prod_{g=0}^m(K+g)}+$$ where the second line emerges from the cancellation of the $Kx^n$ terms in the numerator.
In a similar vein, evaluating the derivative term multiplied by $x$, i.e. $xh'(x)$, leads to $$xh'(x)=\frac{x}{(K+1)K\Gamma(K)}+\frac{2x^2}{(K+2)(K+1)K\Gamma(K)}+\cdots+\frac{mx^m}{\Gamma(K)\prod_{g=0}^m(K+g)}+$$ Subtracting the above two equations results in the cancellation of all terms except for the first term of $(x-K)h(x)$ $$xh'(x)-(x-K)h(x)=\frac{1}{\Gamma(K)}\\\Rightarrow xh'(x)=\frac{1}{\Gamma(K)}+(x-K)h(x)$$