Properties of $\sum _{l=0}^{\infty } \left( -1 \right) ^{l}{m+l-1\choose l} \left( 2 \,{\frac {l}{m}}+1 \right) ^{1-\nu}$

Question

The sums defined by:

$$ S(m,\nu):=\sum _{l=0}^{\infty } \left( -1 \right) ^{l}{m+l-1\choose l} \left( {\frac {2 \,l}{m}}+1 \right) ^{1-\nu} $$ are divergent. However they can be Abel-regularised as

$$ s(m,\nu):=\lim _{x\rightarrow 1^{-}}\sum _{l=0}^{\infty } \left( -1 \right) ^{l} {m+l-1\choose l} \left( {\frac {2\,l}{m}}+1 \right) ^{1-\nu}{x}^{l} $$

Numerical and graphical evidence (see figure below) suggest that for all integer $m\geq 1$ and $\forall\nu\in\mathbb{R}^{+}$, we have the following properties:

$$s(m,\nu)\geq 0,$$ $$s(m,\nu)\geq s(m+1,\nu),$$ and $$\lim_{\nu\rightarrow\infty}s(m,\nu)=1.$$

Apart from the obvious case, where $\nu=1$, I have absolutely no idea on how to prove this. Are these relations known or does anyone know how to prove it? Any reference or hints are welcome.

Thank you very much in advance.

Answer 1

The claim regarding $\nu\to\infty$ holds trivially, since $S(m,\nu)$ is absolutely convergent if $\nu>m+1$, and then the limit can be taken termwise (decrease in absolute value w.r.t. $\nu$ $\implies$ DCT).

As for the rest, the following integral representation(s) of $s(m,\nu)$ may be useful.

If $\nu>1$ (or even $\Re\nu>1$) then, using $$\int_0^\infty t^{a-1}e^{-bt}\,dt=\Gamma(a)b^{-a}\qquad(\Re a,\Re b>0)\tag{1}$$ with $a=\nu-1$ and $b=1+2l/m$, and the binomial series $$\sum_{l=0}^\infty\binom{m+l-1}{l}z^l=(1-z)^{-m},\qquad(|z|<1)\tag{2}$$ we obtain

\begin{align*} s(m,\nu)&=\lim_{x\to1^-}\sum_{l=0}^\infty(-x)^l\binom{m+l-1}{l}\left(1+\frac{2l}{m}\right)^{1-\nu} \\\color{gray}{(1)}\quad&=\lim_{x\to1^-}\sum_{l=0}^\infty(-x)^l\binom{m+l-1}{l}\frac1{\Gamma(\nu-1)}\int_0^\infty t^{\nu-2}e^{-(1+2l/m)t}\,dt \\&=\frac1{\Gamma(\nu-1)}\lim_{x\to1^-}\int_0^\infty t^{\nu-2}e^{-t}\sum_{l=0}^\infty\binom{m+l-1}{l}(-xe^{-2t/m})^l\,dt \\\color{gray}{(2)}\quad&=\frac1{\Gamma(\nu-1)}\lim_{x\to1^-}\int_0^\infty t^{\nu-2}e^{-t}(1+xe^{-2t/m})^{-m}\,dt \\&=\frac1{\Gamma(\nu-1)}\int_0^\infty\frac{t^{\nu-2}\,dt}{(e^{t/m}+e^{-t/m})^m}. \end{align*}

I would try this approach to show the decrease of $m\mapsto s(m,\nu)$. [Looks pretty good, and allows an easy asymptotic analysis of $m\to\infty$ by the way.] Also, integration by parts gives $$s(m,\nu)=\frac{1}{\Gamma(\nu)}\int_0^\infty t^{\nu-1}\frac{e^{t/m}-e^{-t/m}}{(e^{t/m}+e^{-t/m})^{m+1}}\,dt,$$ applicable for $\nu>0$ (or $\Re\nu>0$).

Answer 2

I do not know how this could help you. In any manner, it is not a proof.

If $\nu$ is a positive integer $$T_{m,\nu}=\sum _{l=0}^{\infty } (-1)^l\,{m+l-1\choose l} \,\left( {\frac {2l}{m}}+1 \right) ^{1-\nu}$$ is an hypergeometric function $$T_{m,2}=\, _2F_1\left(\frac{m}{2},m;\frac{m+2}{2};-1\right)$$ $$T_{m,3}=\, _3F_2\left(\frac{m}{2},\frac{m}{2},m;\frac{m+2}{2},\frac{m+2}{2};-1\right)$$ $$T_{m,4}=\, _4F_3\left(\frac{m}{2},\frac{m}{2},\frac{m}{2},m;\frac{m+2}{2},\frac{m+2}{2}, \frac{m+2}{2};-1\right)$$ and you can see the general pattern.

The fact that $T_{m,\nu} >0 ~~\forall m,n$ is clear from definition.

Beside what you found $$T_{m,\nu}-T_{m+1,\nu} >0 $$ there is also $$T_{m,\nu+1}-T_{m,\nu} >0 $$ For example, using $\nu=2$ and various values of $m$, we have $$\left( \begin{array}{ccc} m & T_3-T_2 & \text{numerical value} \\ 1 & C-\frac{\pi }{4} & 0.130567 \\ 2 & -\frac{1}{2}+\frac{\log (4)}{2} & 0.193147 \\ 3 & -\frac{9}{16}+\frac{9 C}{8}-\frac{3 \pi }{32} & 0.173437 \\ 4 & -\frac{1}{3}+\frac{\log (4)}{3} & 0.128765 \\ 5 & -\frac{275}{768}+\frac{75 C}{128}-\frac{15 \pi }{512} & 0.086587 \\ 6 & -\frac{49}{320}+\frac{3 \log (4)}{20} & 0.054819 \end{array} \right)$$ This has been checked for a few values of $\nu$.

Now, using $m=10$ and increasing $\nu$ $$\left( \begin{array}{cc} \nu & \text{numerical value} \\ 20 & 0.757273699599990 \\ 30 & 0.952389215988479 \\ 40 & 0.991943127308829 \\ 50 & 0.998685163156552 \\ 60 & 0.999787167124773 \\ 70 & 0.999965609645380 \\ 80 & 0.999994445189100 \\ 90 & 0.999999102847086 \\ 100 & 0.999999855104106 \end{array} \right)$$

Edit

Working with fixed values of $m$, the only explicit expressions I have been able to get are $$T_{1,\nu}=2^{2-2 \nu } \left(\zeta \left(\nu -1,\frac{1}{4}\right)-\zeta \left(\nu-1,\frac{3}{4}\right)\right)$$ $$T_{2,\nu}=2^{-\nu } \left(2^{\nu }-8\right) \zeta (\nu -2)$$