Convergence/Divergence of summation

Question

Is it possible to determine if the following sum converges as $k$ tends to infinity? Wolfram Alpha gave a version in terms of the incomplete gamma function however I didn't see how that could help me. $$\sum_{n=0}^{k}\frac{k!\left(-1\right)^{n}}{x^{n}\left(k-n\right)!}$$

Answer 1

Call your sum $f_k(x)$. The exponential generating function of this is

$$ \eqalign{g(x,t) &= \sum_{k=0}^\infty f_k(x) t^k/k!\cr &= \sum_{n=0}^\infty \sum_{k=n}^\infty \frac{(-1)^n}{x^n (k-n)!} t^k \cr &= \sum_{n=0}^\infty \frac{(-1)^n t^n e^t}{x^n} \cr &= \frac{x e^t}{t+x} }$$ converging if $|t| < |x|$. This has a pole at $t = -x$. But if $f_k(x)$ converged as $k \to \infty$, $g(x,t)$ would converge for all $t$, with no pole. So it must diverge, in fact it must be unbounded.