# Gamma & Zeta Summation $\sum_{n=0}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)}{(n+1)!}=0$

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

• 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 $$\sum_{n=0}^\infty\frac{\zeta(n+s)\Gamma(n+s)}{(n+1)!} = \lim_{N\to\infty}\int_0^\infty \left[\sum_{n=1}^{N}\frac{x^{n}}{n!}\right] \cdot \frac{x^{s-2}}{e^{x}-1}{\rm d}x$$ – Winther Sep 8 '16 at 21:02
• @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 $\mathbb{R}^+$. – Jack D'Aurizio Sep 8 '16 at 21:04
• @Hazem Orabi: The value range for $\Re(s)$ is false. – user90369 Sep 8 '16 at 21:15
• @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. – Winther Sep 8 '16 at 21:51
• @Winther: Starting by partitioning the validity domain is correct as will as your formula, but what next. – Hazem Orabi Sep 10 '16 at 20:30

## 3 Answers

• 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

• 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) – Hazem Orabi Sep 9 '16 at 15:43
• @HazemOrabi are you joking ? so your comment makes no sense as my (long) answer is exactly about PROVING (and not assuming) everything works well – reuns Sep 10 '16 at 10:31

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.

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 $x=1/2.$

• what do you think of the 2 complex variables analytic continuation problem ? – reuns Sep 9 '16 at 2:21
• and you replaced $n!$ by $(n+1)!$ so you lose the $(1-x)^{1-s}$ term I find and replaced $a-1$ by $a$ – reuns Sep 9 '16 at 2:50
• No, I shifted the summation index, which is why R is defined as it is. – SKBMoore Sep 9 '16 at 5:02
• 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) – reuns Sep 9 '16 at 5:12