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I find two interesting limits : \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*} But I have no idea about the second. The process that I investigated it produces a problem : $$\underset{x\geqslant 0}{\sup}\left|\sum_{n=0}^{\infty}\frac{\left(-x\right)^n}{\Gamma(1+\alpha n)}\right|=\begin{cases} 1& 0<\alpha\leqslant 2\\ +\infty& \alpha>2 \end{cases}.$$ But I have no way to prove it. Thanks for any help.

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    $\begingroup$ @LSpice Thanks. $\endgroup$ – Roc Yeats Apr 26 '20 at 23:11
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Your original limits can be generalized: for $0<\alpha<\beta$, if we let $$G(\alpha,\beta):=\sum_{n=0}^\infty(-1)^n\frac{\Gamma(1+n\alpha)}{\Gamma(1+n\beta)},$$ then $\lim\limits_{\alpha\to\beta^-}G(\alpha,\beta)$ and $\lim\limits_{\beta\to\alpha^+}G(\alpha,\beta)$ are both equal to $1/2$ (see the last section).


For the "produced" problem, we have the following integral representation: $$F_\alpha(x):=\sum_{n=0}^\infty\frac{(-x^\alpha)^n}{\Gamma(1+n\alpha)}=\frac{1}{2\pi\mathrm{i}}\int_\lambda\frac{e^{xz}\,dz}{z(1+z^{-\alpha})},$$ where the contour $\lambda$ encircles the negative real axis and the poles of the integrand (we assume that the principal value of $z^{-\alpha}$ is taken here). This is shown using the geometric series for $(1+z^{-\alpha})^{-1}$, with $\lambda$ deformed if needed for $\Sigma\leftrightarrow\smallint$, and Hankel's contour integral for $1/\Gamma$.

Using residues, we can shrink $\lambda$ to encircle the negative real axis closely (or even evaluate $F_\alpha$ completely if $\alpha$ is an integer). For $\alpha>2$, the residues bring exponentials of the form $e^{cx}$ with $c>0$, which imply the unboundedness of $F_\alpha$. For $0<\alpha<1$, the integrand has no poles, and we obtain (in the limit of "closely") $$F_\alpha(x)=I_\alpha(x):=\frac{\sin\alpha\pi}{\alpha\pi}\int_0^\infty\frac{\exp(-xt^{1/\alpha})\,dt}{1+2t\cos\alpha\pi+t^2},$$ positive and decreasing in $x$. For $1<\alpha<2$, there are poles at $z=\exp(\pm\mathrm{i}\pi/\alpha)$, so this time $$F_\alpha(x)=I_\alpha(x)+\frac{2}{\alpha}\exp\left(x\cos\frac\pi\alpha\right)\cos\left(x\sin\frac\pi\alpha\right),$$ which makes a proof of $|F_\alpha(x)|\leqslant 1$ tedious this way (but still possible I believe).


The same approach applies to our "generalized limits". We have $$G(\alpha,\beta)=\frac{1}{2\pi\mathrm{i}}\int_0^\infty e^{-t}\int_\lambda\frac{\exp(t^{\alpha/\beta}z)}{z(1+z^{-\beta})}\,dz\,dt$$ and, if we deform $\lambda$ to fit in $\Re z<1$ strictly, then both $\alpha\to\beta^-$ and $\beta\to\alpha^+$ can be taken under the integrals, and moreover, the integrations can be interchanged, resulting in $$\frac{1}{2\pi\mathrm{i}}\int_\lambda\frac{dz}{z(1-z)(1+z^{-\alpha})}=-\operatorname*{Res}_{z=1}\ldots=\frac12$$ (by completing the part of $\lambda$ inside $|z|=R$ with the larger arc, and taking $R\to\infty$).

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  • $\begingroup$ I'm still looking for a simple way to show $|F_\alpha(x)|\leqslant1$ for $1<\alpha<2$. The above shows that $F_\alpha$ is a difference of two functions, an exponentially decaying oscillating one, and a decreasing tending-to-zero one. Perhaps an alternative integral representation would do, or even the original contour integral could be modified suitably. $\endgroup$ – metamorphy Aug 24 '20 at 15:35

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