Generating function of the Confluent Hypergeometric Function of the First Kind

Question

Let $x>0$ and $t\in(-1,1)$. Consider

$$\sum_{m=1}^\infty t^m \sum_{l=1}^m\binom{m-1}{l-1}\frac{(-x)^l}{l!}\,.$$

Can you find a closed expression for this series? Thank you for your time!

It reminds me the generating function associated with the generalized Laguerre polynomials $L_{m}^{(\alpha)}(x)$ with $\alpha=-1$. In addition, maybe it can be useful to note that the series can be expressed as $$-x \sum_{m=1}^\infty t^m\, {}_1\!F(1-m;2;x)\,,$$ where I have introduced the "Confluent Hypergeometric Function of the First Kind" ${}_1\!F(a;b;c)$.

Answer 1

For $0 < t < 1/2$ it seems to be equal to $$ \boxed{e^{xt/(t-1)} - 1}. $$ The proof is a bit tedious, maybe someone will propose something better later. Consider $$ S = \sum_{m=1}^{\infty} t^m \sum_{l=1}^m \binom{m-1}{l-1} \frac{(-x)^l}{l!}. \tag{1} $$ I will use integral representation of binomial coefficient, i.e. $$ {n\choose k} = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{(1+z)^n}{z^{k+1}} \; dz. \tag{IR} $$ Taking (IR) into account (1) transforms into $$ S = \sum_{m=1}^{\infty} t^m \sum_{l=1}^m \frac{1}{2\pi i} \oint_{|z|=1} \frac{(1+z)^{m-1}}{z^l} \; dz \frac{(-x)^l}{l!}, \tag{2} $$ as an integration contour I choose a circle radius $1$, counterclockwise. Now we may note the following "overflow equality" $$ l > m \rightarrow \oint_{|z|=1} \frac{(1+z)^{m-1}}{z^l} \; dz = 0. \tag{OE} $$ With (OE) we can extend the sum in (2) to infinity $\sum_{l=1}^m \to \sum_{l=1}^\infty$ and move everything into the integral $$ S = \frac{1}{2\pi i} \oint_{|z|=1} \underbrace{\sum_{m=1}^{\infty} t^m (1+z)^{m-1}}_{t/(1-t-tz)} \underbrace{\sum_{l=1}^\infty \frac{(-x)^l}{z^l l!}}_{e^{-x/z} - 1}\; dz, \tag{3} $$ which yields after simplification $$ S = \frac{1}{2\pi i}\oint_{|z|=1} \frac{e^{-x/z} - 1}{(1-t)/t-z} \; dz. \tag{4} $$ For convenience I perform variables change in (4) $z \to 1/\xi$, one may note that integration contour still remains unit circle,but the direction is clockwise now $$ S = \frac{t}{2\pi i (1-t)}\oint_{|\xi|=1} \frac{e^{-x\xi} - 1}{\xi(\xi- t/(1-t))}\; d\xi. \tag{5} $$ For $0 < t < 1/2$ this function has $2$ poles inside the integration contour ($\xi = 0$, $\xi = t/(1-t)$), using residues we get the result $$ S = e^{xt/(t-1)} - 1. \tag{R} $$ As a simple test one can use Wolfram Mathematica

fnum[t_, x_] := 
 NSum[t^m*Sum[Binomial[m - 1, l - 1]*(-x)^l/l!, {l, 1, m}], {m, 1, 
   1000}]
fan[t_, x_] = Exp[t*x/(t - 1)] - 1

Comparison for $t= 0.1,0.2,0.3,0.4$ follows

Answer 2

In addition to the excellent answer by @guest, it is also possible to use the generating function for the generalized Laguerre polynomials \begin{equation} (1-t)^{-\alpha-1}\exp\left(\frac{xt}{t-1}\right)=\sum_{n=0}^{\infty}L^{(\alpha% )}_{n}\left(x\right)t^{n} \text{ when } |t|<1 \end{equation} as guessed in the OP. From the representation of the series in terms of confluent hypergeometric function \begin{equation} S(t,x)=-x \sum_{m=1}^\infty t^m\, {}_1F_1(1-m;2;x) \end{equation} and using the generalized Laguerre polynomials representation \begin{equation} L^{(\alpha)}_{n}\left(x\right)=\frac{{\left(\alpha+1\right)_{ n}}}{n!}{{}_{1}F_{1}}\left({-n\atop\alpha+1};x\right) \end{equation} with $\alpha=1,n=m-1$, it comes \begin{align} S(t,x)&=-x \sum_{m=1}^\infty t^m\frac{(m-1)!}{{\left(2\right)_{ m-1}}}L_{m-1}^{(1)}(x)\\ &=-x \sum_{m=1}^\infty \frac{t^m}{m}L_{m-1}^{(1)}(x)\\ \end{align} Then, using the generating function above \begin{align} \frac{\partial S(t,x)}{\partial t}&=-x\sum_{m=1}^\infty t^{m-1}L_{m-1}^{(1)}(x)\\ &=-x\frac{\exp\left(\frac{xt}{t-1}\right)}{(1-t)^2}\\ &=\frac{\partial}{\partial t}\left[ \exp\left(\frac{xt}{t-1}\right)\right] \end{align} With $S(0,x)=0$ we have \begin{align} S&=\int_0^t \frac{\partial S(\tau,x)}{\partial \tau}\,d\tau\\ &= \exp\left(\frac{xt}{t-1}\right)-1 \end{align}