Is there a convergence for the series $ \sum_{i=0}^{\infty} \frac{(x-i)^i}{i!} $ The following series converges to exponential.
$\sum_{i=0}^{\infty} \frac{x^i}{i!} = e^{x}$ 
Is the convergence of the following series known?
$\sum_{i=0}^{\infty} \frac{(x-i)^i}{i!}$
 A: This series diverges. In fact the terms $\frac{(x-i)^i}{i!}$ do not even tend to zero for any fixed $x$.
A: Your series is asymptotically equivalent to 
$$
\sum_{i=0}^\infty\frac{(-1)^ii^i}{i!}
$$
Absolute value of each term of the above series is greater than  $1$ (except the first two terms) and hence it diverges and so does the original series.
A: $$a_n=\frac{(x-n)^n}{n!}\implies A=\frac{a_{n+1}}{a_n}=\frac1{n+1 }\frac{ (x-n-1)^{n+1}} {(x-n)^n }=\frac{-1}{n+1 }\frac{ (n+1-x)^{n+1}} {(n-x)^n }$$ $$B=\frac{ (n+1-x)^{n+1}} {(n-x)^n }\implies\log(B)=(n+1)\log(n+1-x)-n\log(n-x)$$ 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 $$\log(B)=1+\log \left(n\right)+\frac{1}{2
   n}+\frac{\frac{x^2}{2}-\frac{1}{6}}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$B=e^{\log(B)}=e n+\frac{e}{2}+\frac{e \left(12 x^2-1\right)}{24
   n}+O\left(\frac{1}{n^2}\right)$$ $$A=\frac{-B}{n+1}=-e+\frac{e}{2 n}-\frac{e \left(12 x^2+11\right)}{24
   n^2}+O\left(\frac{1}{n ^3}\right)$$
