Checking Jaynes' formula 6.108 for $\sum\limits_{m=0}^\infty{m+a \choose m} m^nx^m$

Question

I read about this equation, $$\sum_{m=0}^\infty{m+a \choose m} m^nx^m = (x\frac{d}{dx})^n\frac{1}{(1-x)^{a+1}} \;\;\;(|x| < 1)\;\;\; (6.108)$$ As far as i know, $$\sum_{m=0}^\infty{m+a \choose m}x^m = \frac{1}{(1-x)^{a+1}}, \; \; \; (|x| <1 ). $$ This can be proved by the Taylor series which expands $\frac{1}{(1-x)^{a+1}} \;$ at x = 0. And the (6.108) equation may be an error. Help, Thanks.

Answer 1

I haven't checked the details fully, but it looks reasonable to me. For the sake of intuition, let's assume that we can move the derivative through the infinite sum. Take the second expression and differentiate it once wrt $x$, then multiply by $x$. This gives a factor $m$ and decreases the power of $x$ by one, but then increases it back when multiplying by $x$. More precisely, $$ x \cdot \frac{d}{dx} x^m = x \cdot m x^{m-1} = m x^m. $$ So, informally, we see that the 'operator' $x (d/dx)$ when applied to $x^m$ acts as scalar multiplying by $m$. Hence if you do this $m$ times to each term in the series, the you get the desired result.

Hopefully that clears up any issues! :)