4
$\begingroup$

According to Gamma Summation & Zeta Summation: $$ \sum_{n=0}^{\infty} {(-1)^n \frac{\Gamma(n+s) \zeta(n+s)}{(n+1)!}}=\Gamma(s-1) \quad : \space Re\{s\}<2 $$

Show that: $$ \sum_{n=0}^{\infty} {\frac{\Gamma(n+s) \zeta(n+s)}{(n+1)!}}=0 \quad : \space Re\{s\}<1 $$

In other words, the Even & Odd parts are convergent series, equaling sums, and different signs: $$ \sum_{n=0}^{\infty} {\frac{\Gamma(2n+s) \zeta(2n+s)}{(2n+1)!}}=-\sum_{n=0}^{\infty} {\frac{\Gamma(2n+1+s) \zeta(2n+1+s)}{(2n+1+1)!}}=\frac{\Gamma(s-1)}{2} \space : \space Re\{s\}<1 $$

$\endgroup$
  • $\begingroup$ From the relation $\zeta(s)\Gamma(s) = \int_0^\infty \frac{x^{s-1}}{e^x-1}{\rm d}x$ it follows that we should have something like $$\sum_{n=0}^\infty\frac{\zeta(n+s)\Gamma(n+s)}{(n+1)!} = \lim_{N\to\infty}\int_0^\infty \left[\sum_{n=1}^{N}\frac{x^{n}}{n!}\right] \cdot \frac{x^{s-2}}{e^{x}-1}{\rm d}x$$ $\endgroup$ – Winther Sep 8 '16 at 21:02
  • 1
    $\begingroup$ @Winther: it is not that straighforward since the above integral representation for $\Gamma(s)\zeta(s)$ only holds for $\text{Re}(s)>1$, and $x^{s-2}$ is never integrable over $\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio Sep 8 '16 at 21:04
  • $\begingroup$ @Hazem Orabi: The value range for $\Re(s)$ is false. $\endgroup$ – user90369 Sep 8 '16 at 21:15
  • 1
    $\begingroup$ @JackD'Aurizio Yeah a bit of care is needed, but I think one can use this it at least for some special cases. If for example we only consider $\Re s \in (0,1)$ then $\Re(n+s) > 1$ for $n>0$ so the integral rep. holds for all but the first term which gives $$\sum_{n=0}^\infty\frac{\zeta(n+s)\Gamma(n+s)}{(n+1)!} = \zeta(s)\Gamma(s) + \int_0^\infty \frac{e^x-1-x}{e^x-1}x^{s-2}{\rm d}x$$ Don't see right now how to show that this evaluates to $0$, but it does seem to hold numerically. $\endgroup$ – Winther Sep 8 '16 at 21:51
  • $\begingroup$ @Winther: Starting by partitioning the validity domain is correct as will as your formula, but what next. $\endgroup$ – Hazem Orabi Sep 10 '16 at 20:30
1
$\begingroup$
  • note that $\frac{\Gamma(s+n)}{(n+1)!} = \frac{\Gamma(n+s)}{\Gamma(n+2)} = \mathcal{O}(n^{s-2})$, and since $\zeta(s+n) \to 1$, it converges for $Re(s) < 1$

  • but since $$\Gamma(s) \zeta(s) = \int_0^\infty \frac{x^{s-1} }{e^x-1}dx, \qquad \color{red}{\text{only for }} Re(s) > 1$$

    (for proving it : $\int_0^\infty x^{s-1} e^{-nx} dx = n^{-s}\Gamma(s), Re(s) > 0$ and $\frac{1}{e^x-1} = \sum_{n=1}^\infty e^{-nx}, x > 0$)

  • we have to consider the following regularized version $$\begin{eqnarray}\sum_{n=0}^\infty z^n \frac{\Gamma(s+n)\zeta(s+n)}{(n+1)!} &=& \int_0^\infty \sum_{n=1}^\infty z^n \frac{x^n}{n!}\frac{x^{s-2} }{e^x-1}dx \\ &=& \int_0^\infty (e^{zx}-1)\frac{x^{s-2} }{e^x-1}dx \\ & =& -\zeta(s-1)\Gamma(s-1)+ \int_0^\infty x^{s-2}\sum_{n=1}^\infty e^{-(n-z)x}dx \\ &=& \Gamma(s-1)(-\zeta(s-1)+\sum_{n=1}^\infty (n-z)^{1-s}) \\ &=& \Gamma(s-1) ((1-z)^{1-s}+\zeta(s-1,1-z)-\zeta(s-1)) \\ & &\qquad\qquad\qquad \color{red}{(Re(s) > 2, |z| < 1)}\end{eqnarray}$$ where $\zeta(s,a) = \sum_{n=1}^\infty (n+a)^{-s}$ is the Hurwitz zeta function, having a nice analytic continuation too, such that $\zeta(s) = \lim_{a\to 0} \zeta(s,a)$ also for $Re(s) < 1$. (**)

Hence, assuming the analytic continuation (*) of two variables $z,s$ works well (keeping $|z| < 1$) this stays true for $Re(s) < 1$ where we can extend to $|z| = 1$ and get $$\sum_{n=0}^\infty \frac{\Gamma(s+n)\zeta(s+n)}{(n+1)!} = \lim_{z \to 1^-}\sum_{n=0}^\infty z^n \frac{\Gamma(s+n)\zeta(s+n)}{(n+1)!} $$ $$\qquad =\lim_{z \to 1^-} \Gamma(s-1) ((1-z)^{1-s}+\zeta(s-1,1-z)-\zeta(s-1)) = 0\quad (Re(s) < 1, -s \not \in \mathbb{N})$$

the $\lim_{z \to 1^-} $ is justified by the fact the LHS converges absolutely when $|z| \le 1, Re(s) < 1$ (so it is continuous in $z$), and the fact the RHS is continuous in $|z| \le 1$ too

(*) and it does since the RHS $\Gamma(s-1) ((1-z)^{1-s}+\zeta(s-1,1-z)-\zeta(s-1))$ is obviously holomorphic/analytic/meromorphic in $s$, while the LHS is also holomorphic in $s$, by showing $\sum_{n=0}^\infty z^n \frac{\frac{d}{ds}\Gamma(s+n)\zeta(s+n)}{(n+1)!}$ converges absolutely when $|z| < 1$, and when $|z|= 1, Re(s) < 1$

(**) integrating by parts $\zeta(s,a) = \sum_{n=1}^\infty (n+a)^{-s} = \int_{a-\epsilon}^\infty \sum_{n=1}^\infty \delta(x-n-a)x^{-s}dx = s \int_a^\infty \lfloor x+a \rfloor x^{-s-1} dx$ $ $ $= \frac{s a^{-s}}{s-1}+a^{-s}+s \int_a^\infty (\lfloor x+a \rfloor-x-a) x^{-s-1} dx \ \ (Re(s) > 0)$, and we can integrate by parts several times the same way

$\endgroup$
  • 1
    $\begingroup$ Assuming the analytic continuation of the two variables (*) (specially for $Re\{s\}<2$) which lead to the absolute convergence of the sum (for $Re\{s\}<1 \space\&\space |z|\le1$) is not solid. As well as, the limit should exist in both directions, because your assumption is reflecting the analytic continuation for Hurwitz zeta too. (Edmund Landau: Handbuch der Lehre von der Verteilung der Primzahlen) $\endgroup$ – Hazem Orabi Sep 9 '16 at 15:43
  • $\begingroup$ @HazemOrabi are you joking ? so your comment makes no sense as my (long) answer is exactly about PROVING (and not assuming) everything works well $\endgroup$ – reuns Sep 10 '16 at 10:31
1
$\begingroup$

In Handbuch, Edmund Landau provided the identity of rising factorial: $$ \begin{align} \,& \zeta(s)=\frac{s}{s-1}-\sum _{n=1}^{\infty}\left[\frac{s(s+1)\cdots(s+n-1)}{(n+1)!}\left(\zeta(s+n)-1\right)\right] \,\,\,\colon\,s\in\mathbb{C} \\[2mm] \,& \Rightarrow\,\Gamma(s)\zeta(s)=\frac{\Gamma(s+1)}{s-1}-\sum _{n=1}^{\infty}\left[\frac{\Gamma(s+n)}{(n+1)!}\left(\zeta(s+n)-1\right)\right] \\[2mm] \,& \qquad\qquad\quad\,\,\,=\Gamma(s-1)+\Gamma(s)-\sum _{n=1}^{\infty}\left[\frac{\Gamma(s+n)\zeta(s+n)}{(n+1)!}-\frac{\Gamma(s+n)}{(n+1)!}\right]\,\Rightarrow \end{align} $$

$$ \sum _{n=0}^{\infty}\frac{\Gamma(s+n)\zeta(s+n)}{(n+1)!}-\sum _{n=0}^{\infty}\frac{\Gamma(s+n)}{(n+1)!}=\Gamma(s-1) \quad\colon\,s\in\mathbb{C}\qquad\qquad\qquad\tag{1} $$

And from the summation identity of gamma function: $$ \begin{align} \,& \quad\sum_{n=0}^{\infty}\frac{\Gamma(n+s)}{n!}=0 \,\,\colon\,Re\{s\}\lt0 \,\,\Rightarrow\,\, \sum_{n=0}^{\infty}\frac{\Gamma(n+s-1)}{n!}=0 \,\,\colon\,Re\{s\}\lt1\,\,\Rightarrow \\[2mm] \,& -\sum_{n=\color{red}{1}}^{\infty}\frac{\Gamma(n+s-1)}{n!}=\Gamma(s-1) \,\Rightarrow\, -\sum_{n=\color{red}{0}}^{\infty}\frac{\Gamma(n+s)}{(n+1)!}=\Gamma(s-1) \,\,\colon\,Re\{s\}\lt1\tag{2} \\[2mm] \end{align} $$ Limiting the range of $(1)$ to be $\small\left\{\,Re\{s\}\lt1\,\right\}$ like $(2)$,
the gamma series from LHS of $(1)$ will cancel $\small\Gamma(s-1)$ of RHS,
leaving: $$ \sum_{n=0}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)}{(n+1)!}=0\,\,\,\colon\space Re\{s\}\lt1 $$ As desired.

$\endgroup$
-1
$\begingroup$

The proposer wants $R(1,s-1)$ of the function defined below: $$R(x,a)=\sum_{n=1}^\infty \dfrac{x^n}{n!}\Gamma(n+a)\,\zeta(n+a).$$ The $\Gamma\cdot\zeta$ product has a well-known integral relationship, as alluded to by Jack D'Aurizio. Switch integration and summation $-$ the summation is an exponential $-$ and one gets $$R(x,a)=\int_0^\infty du \,u^{a-1} \,\dfrac{e^{x\,u}-1}{e^{u}-1 }=\Gamma(a)(\zeta(a,1-x)-\zeta(a)). $$ (Need $a>0$ for the integral to converge, but the right-hand side is an analytic continuation.)
The Hurwitz $\zeta$ reduces to the Riemann $\zeta$ for $x=1,$ and the formula is proved. Incidentally, the Hurwitz $\zeta$ also has closed forms in terms of the Riemann $\zeta$ for $x=-1,$ as the proposer has noted, but also for $x=1/2.$

$\endgroup$
  • $\begingroup$ what do you think of the 2 complex variables analytic continuation problem ? $\endgroup$ – reuns Sep 9 '16 at 2:21
  • $\begingroup$ and you replaced $n!$ by $(n+1)!$ so you lose the $(1-x)^{1-s}$ term I find and replaced $a-1$ by $a$ $\endgroup$ – reuns Sep 9 '16 at 2:50
  • $\begingroup$ No, I shifted the summation index, which is why R is defined as it is. $\endgroup$ – SKBMoore Sep 9 '16 at 5:02
  • $\begingroup$ if you meant $\zeta(a,1-x) = \sum_{n=0}^\infty (n+1-x)^{-a}$ then you need to talk about the $(1-x)^{-a}$ term (anyway you need to explain the analytic continuation process, as i did) $\endgroup$ – reuns Sep 9 '16 at 5:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.