Does $\sum_{k=0}^{\infty}\frac{1}{(k+1)!}\left(\prod_{l=0}^{k}\left(1-ls\right)\right) x^{k+1}$ converge?

Question

Does

$$\sum_{k=0}^{\infty}\frac{1}{(k+1)!}\left(\prod_{l=0}^{k}\left(1-ls\right)\right) x^{k+1}$$

converge for $s\in(0,1)$? If so, to what value?

Answer 1

In fact, if $ s\neq 0 $ then the sum will converge if and only if $\left|sx\right|<1 $, and if $ s=0 $ then it will converge for any real $ x \cdot $

Let $ \left(s,x\right)\in\mathbb{R}^{*}\times\mathbb{R} $, such that $ sx\in\left]-1,1\right[ $, Doing a simple change of index, gives : \begin{aligned} \sum_{k=0}^{+\infty}{\frac{1}{\left(k+1\right)!}\prod_{\ell =0}^{k}{\left(1-\ell s\right)}x^{k+1}}=\sum_{k=1}^{+\infty}{\frac{1}{k!}\prod_{\ell =0}^{k-1}{\left(1-\ell s\right)}x^{k}}&=\sum_{k=1}^{+\infty}{\frac{1}{k!}\prod_{\ell =0}^{k-1}{\left(\frac{1}{s}-\ell\right)}\left(sx\right)^{k}}\\&=\sum_{k=1}^{+\infty}{\binom{\frac{1}{s}}{k}\left(sx\right)^{k}} \end{aligned}

Where $ \binom{\alpha}{n} $ designates, for any $ \left(\alpha,n\right)\in\mathbb{C}\times\mathbb{N} $, the product $ \frac{\alpha\left(\alpha -1\right)\cdots\left(\alpha -n+1\right)}{n!} \cdot $

The final result would, then, be : $$ \left(1+sx\right)^{\frac{1}{s}}-1 $$

If $ s=0 $, then, as we said, the sum will converge for all $ x\in\mathbb{R} $ to : $$ \exp{x}-1 $$