How can I prove $\lim_{s\to 1^-}\sum_{n=0}^{\infty}\left(-1\right)^n\frac{\Gamma(1+ns)}{\Gamma(1+n)}=\frac{1}{2}$

Question

I find two interesting limits (related post on MO): \begin{align*} \frac{1}{2}& =\lim_{s\to 1^-}\sum_{n=0}^{\infty}\left(-1\right)^n\frac{\Gamma(1+ns)}{\Gamma(1+n)}\\ & =\lim_{s\to 1^+}\sum_{n=0}^{\infty}\left(-1\right)^n\frac{\Gamma(1+n)}{\Gamma(1+ns)}. \end{align*} It's definition of $1-1+1-1+\cdots$ being Le Roy summable.

Further conjecture is $$\underset{x\geqslant 0}{\sup}\left|\sum_{n=0}^{\infty}\frac{\left(-x\right)^n}{\Gamma(1+\alpha n)}\right|=\begin{cases} 1& 0<\alpha\leqslant 2\\ +\infty& \alpha>2 \end{cases}.$$ If this conjecture is valid, I can apply Dominated Convergence Theorem to calculate the limits above.

I just used the Corollary 3.7 of Mittag-Leffler Functions, Related Topics and Applications to prove $$\underset{x\geqslant 0}{\sup}\left|\sum_{n=0}^{\infty}\frac{\left(-x\right)^n}{\Gamma(1+\alpha n)}\right|=\begin{cases} \mathrm{finite}& 0<\alpha\leqslant 2\\ +\infty& \alpha>2 \end{cases}.$$ However, I have no idea to improve finite to 1. Thanks for any help.

Answer 1

Using the integral representation of the Gamma function we have for $0\le s<1$

$$\begin{align} \sum_{n=0}^\infty \frac{(-1)^n}{n!}\Gamma(1+ns)&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^\infty x^{ns}e^{-x}\,dx\tag1\\\\ &=\int_0^\infty e^{-x}\sum_{n=0}^\infty \frac{(-1)^nx^{ns}}{n!}\,dx\tag2\\\\ &=\int_0^\infty e^{-x}e^{-x^s}\,dx\\\\ \end{align}$$

Letting $s\to 1^-$ reveals

$$\begin{align} \lim_{s\to 1^-}\sum_{n=0}^\infty \frac{(-1)^n}{n!}\Gamma(1+ns)&=\int_0^\infty e^{-2x}\,dx\\\\ &=\frac12 \end{align}$$

as expected!

---

NOTE:

In going from $(1)$ to $(2)$ we noted that $\left|\sum_{n=0}^N \frac{(-1)^nx^{ns}}{n!}\right|\le e^{x^s}$. And $\int_0^\infty e^{-x}e^{x^s}\,dx<\infty$ for $0\le s<1$. Application of the Dominated convergence theorem guarantees that the interchange of the series and integral is legitimate.