I'm reading the 3rd edition of Sakurai and Napolitano's Modern Quantum Mechanics, and have hit a bit of a snag with a sum involving the Hermite polynomials.
The authors define the Hermite polynomials through the generating function
\begin{equation*} g(x,t) = e^{-t^{2}+2xt} = \sum_{n=0}^{\infty}{H_{n}(x)\frac{t^{n}}{n!}} \end{equation*}
and derive the familiar relation that
\begin{equation} H_{n}'(x) = 2nH_{n-1}(x) \end{equation}
(where prime represents differentiation with respect to the argument $x$). Then, (on p. 100), they write
\begin{align} \frac{\partial g}{\partial t} &= -2tg(x,t) + 2xg(x,t)\\ &= -\sum_{n=0}^{\infty}{2H_{n}(x)\frac{t^{n+1}}{n!}} + \sum_{n=0}^{\infty}{2xH_{n}(x)\frac{t^{n}}{n!}}\\ &= -\sum_{n=0}^{\infty}{2nH_{n-1}(x)\frac{t^{n}}{n!}} + \sum_{n=0}^{\infty}{2xH_{n}(x)\frac{t^{n}}{n!}}. \end{align}
Somehow, I'm having trouble seeing how the authors arrive at the ultimate equality. It seems like $H'_{n}(x) = 2nH_{n-1}(x)$ is used, but the derivative with respect to $x$ doesn't show up in the series and I don't see how it could be incorporated by considering the formula as a series expansion of some function.
I'm sure I'm missing something obvious, and would appreciate any advice-I've been hung up on this for a while.