This week's rainy Sunday afternoon problem:
The function $f(x,\alpha)$ for $\alpha$ real and positive, is defined as:
$$ f(x,\alpha) = \sum_{k=0}^{\infty}(-1)^{k}\frac{x^k}{k!^{\alpha}}$$
Find all $\alpha$ such that $\lim_{x\to\infty}f(x,\alpha)$ exists.
For $\alpha>0$, by the Stirling-formula, we have: $$f(x,\alpha)=\sum\limits_{k=0}^{\infty}(-1)^k\dfrac{x^k}{k!^\alpha}\approx C\cdot\sum\limits_{k=1}^{\infty}(-1)^k\dfrac{x^k}{(\frac{k^\alpha}{e^\alpha})^k}=C\cdot\sum\limits_{k=1}^{\infty}(-1)^k(\dfrac{e^\alpha\cdot x}{k^{\alpha}})^k.$$ Thus by Leibniz criteria, for $x>0$, we have conditionally convergence. For $x=0$, $f(0,\alpha)=0$, and for $x<0$, let $y:=(-x)^{1/\alpha}$. For a fixed $x$, there is a positive integer $M$, such that for every $k>M$: $$(\frac{e\cdot y}{k})^\alpha<\frac{1}{2}.$$ Thus we have: $$C\cdot\sum\limits_{k=0}^{\infty}(-1)^k(\dfrac{e^\alpha\cdot x}{k^{\alpha}})^k<constant+C\cdot\sum\limits_{k=M+1}^{\infty}(\dfrac{1}{2})^k<\infty.$$