Confluent hypergeometric function II

Kummer's transformation for ${}_{1}F_{1}$

Here is the proof of this transformation posted by @infseriesbot \[ \sum_{n=0}^{\infty} \frac{z^n}{(a)_{n+1}} = e^{z}\sum_{n=0}^{\infty} \frac{(-z)^n}{(a+n)}\]

Proof:

Recall the definition of the Pochhammer polynomial (rising factorial) \[(a)_{n} = a(a+1)\cdots(a+n-1)\] Hence \begin{align*} S = \sum_{n=0}^{\infty} \frac{z^n}{(a)_{n+1}} =& \sum_{n=0}^{\infty} \frac{z^n}{a(a+1)\cdots(a+n)}\\ =& \sum_{n=0}^{\infty} \frac{z^n}{a\left[(a+1)\cdots(a+n)\right]}\\ =& \sum_{n=0}^{\infty} \frac{z^n}{a(a+1)_{n}}\\ =& \frac{1}{a} \sum_{n=0}^{\infty} \frac{(1)_{n}}{(a+1)_{n}} \frac{z^n}{(1)_{n}}\\ =& \frac{1}{a} \sum_{n=0}^{\infty} \frac{(1)_{n}}{(a+1)_{n}} \frac{z^n}{n!}\\ =& \frac{1}{a} {}_{1}F_{1}\left(1,a+1,z\right)\\ \end{align*} ${}_{1}F_{1}(a,b,z)$ is the confluent hypergeometric function, also denoted $M(a,b,z)$. This function satisfies the Kummer’s Transformation: \[ {}_{1}F_{1}(a,b,z) = e^{z} {}_{1}F_{1}(b-a,b,-z)\] Therefore \begin{align*} \frac{1}{a}{}_{1}F_{1}\left(1,a+1,z\right) =& \frac{e^{z}}{a} {}_{1}F_{1}(a,a+1,-z)\\ =& \frac{e^{z}}{a} \sum_{n=0}^{\infty} \frac{(a)_{n}}{(a+1)_{n}} (-z)^n\\ =& \frac{e^{z}}{a} \sum_{n=0}^{\infty} \frac{a(a+1)\cdots(a+n-1)}{(a+1)(a+2)\cdots(a+n-1)(a+n)} (-z)^n\\ =& \frac{e^{z}}{a}\sum_{n=0}^{\infty} \frac{a(a+1)\cdots(a+n-1)}{(a+1)(a+2)\cdots(a+n-1)(a+n)} (-z)^n\\ =& \frac{e^{z}}{a} \sum_{n=0}^{\infty} \frac{a}{(a+n)} (-z)^n\\ =& e^{z}\sum_{n=0}^{\infty} \frac{(-z)^n}{(a+n)} \end{align*} Hence \[ \boxed{ \sum_{n=0}^{\infty} \frac{z^n}{(a)_{n+1}} = e^{z}\sum_{n=0}^{\infty} \frac{(-z)^n}{(a+n)}}\]