Question: The following series converges to exponential.

$\sum\_{i=0}^{\infty} \frac{x^i}{i!} = e^{x}$ 

Is the convergence of the following series known?

$\sum\_{i=0}^{\infty} \frac{(x-i)^i}{i!}$

Answer: 
This series diverges. In fact the terms $\frac{(x-i)^i}{i!}$ do not even tend to zero for any fixed $x$.
$$a\_n=\frac{(x-n)^n}{n!}\implies A=\frac{a\_{n+1}}{a\_n}=\frac1{n+1 }\frac{ (x-n-1)^{n+1}} {(x-n)^n }=\frac{-1}{n+1 }\frac{ (n+1-x)^{n+1}} {(n-x)^n }$$ $$B=\frac{ (n+1-x)^{n+1}} {(n-x)^n }\implies\log(B)=(n+1)\log(n+1-x)-n\log(n-x)$$ Now, for large values of $n$, use Taylor$$p \log (p-x)=p \log \left(p\right)-x-\frac{x^2}{2 p}-\frac{x^3}{3
 p^2}+O\left(\frac{1}{p^3}\right)$$ Apply to each term and simplify to get $$\log(B)=1+\log \left(n\right)+\frac{1}{2
 n}+\frac{\frac{x^2}{2}-\frac{1}{6}}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$B=e^{\log(B)}=e n+\frac{e}{2}+\frac{e \left(12 x^2-1\right)}{24
 n}+O\left(\frac{1}{n^2}\right)$$ $$A=\frac{-B}{n+1}=-e+\frac{e}{2 n}-\frac{e \left(12 x^2+11\right)}{24
 n^2}+O\left(\frac{1}{n ^3}\right)$$