# On the Sequences of Gamma & Zeta Limits

Considering the Odd & Even Sequences of Gamma & Zeta Limits: \begin{align} & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)(2n+2)}}\space\quad=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-1)}{2}-\frac{\Gamma(x+1)\zeta(x+1)}{2!}\right]\,, \\[2mm] & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)\cdots(2n+3)}}=\lim_{x\rightarrow0}\left[+\frac{\Gamma(x-2)}{2}-\frac{\Gamma(x-1)\zeta(x-1)}{1!}-\frac{\Gamma(x+1)\zeta(x+1)}{3!}\right]\,, \\[2mm] & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)\cdots(2n+4)}}=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-3)}{2}-\frac{\Gamma(x-1)\zeta(x-1)}{2!}-\frac{\Gamma(x+1)\zeta(x+1)}{4!}\right]\,, \\[2mm] & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)\cdots(2n+5)}}=\lim_{x\rightarrow0}\left[+\frac{\Gamma(x-4)}{2}-\frac{\Gamma(x-3)\zeta(x-3)}{1!}-\cdots-\frac{\Gamma(x+1)\zeta(x+1)}{5!}\right]\,, \\[2mm] & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)\cdots(2n+6)}}=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-5)}{2}-\frac{\Gamma(x-3)\zeta(x-3)}{2!}-\cdots-\frac{\Gamma(x+1)\zeta(x+1)}{6!}\right]\,, \quad\cdots \\[2mm] & {\bf=} \space\color{blue}{\frac{1}{2}}-\frac{\gamma}{2}{\bf\,,} \quad\color{blue}{\frac{7}{24}}-\frac{\gamma}{6}+\frac{\zeta'(-1)}{1}{\bf\,,} \quad\color{blue}{\frac{1}{9}}-\frac{\gamma}{24}+\frac{\zeta'(-1)}{2}{\bf\,,} \\[2mm] & \quad\color{blue}{\frac{277}{8640}}-\frac{\gamma}{120}+\frac{\zeta'(-1)}{6}+\frac{\zeta'(-3)}{6}{\bf\,,} \quad\color{blue}{\frac{79}{10800}}-\frac{\gamma}{720}+\frac{\zeta'(-1)}{24}+\frac{\zeta'(-3)}{12}{\bf\,,} \quad\cdots \\[2mm] & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \end{align}

\begin{align} & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)(2n+1)}}\space\quad=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-1)}{2}+\frac{\Gamma(x+0)\zeta(x+0)}{1!}\right]\,, \\[2mm] & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)\cdots(2n+2)}}=\lim_{x\rightarrow0}\left[+\frac{\Gamma(x-2)}{2}+\frac{\Gamma(x+0)\zeta(x+0)}{2!}\right]\,, \\[2mm] & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)\cdots(2n+3)}}=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-3)}{2}+\frac{\Gamma(x-2)\zeta(x-2)}{1!}+\frac{\Gamma(x+0)\zeta(x+0)}{3!}\right]\,, \\[2mm] & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)\cdots(2n+4)}}=\lim_{x\rightarrow0}\left[+\frac{\Gamma(x-4)}{2}+\frac{\Gamma(x-2)\zeta(x-2)}{2!}+\frac{\Gamma(x+0)\zeta(x+0)}{4!}\right]\,, \\[2mm] & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)\cdots(2n+5)}}=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-5)}{2}+\frac{\Gamma(x-4)\zeta(x-4)}{1!}+\cdots+\frac{\Gamma(x+0)\zeta(x+0)}{5!}\right]\,, \quad\cdots \\[2mm] & {\bf=} \space\color{red}{\frac{1}{2}}+\frac{\zeta'(0)}{1}{\bf\,,} \quad\color{red}{\frac{3}{8}}+\frac{\zeta'(0)}{2}{\bf\,,} \quad\color{red}{\frac{11}{72}}+\frac{\zeta'(0)}{6}+\frac{\zeta'(-2)}{2}{\bf\,,} \\[2mm] & \quad\color{red}{\frac{25}{576}}+\frac{\zeta'(0)}{24}+\frac{\zeta'(-2)}{4}{\bf\,,} \quad\color{red}{\frac{137}{14400}}+\frac{\zeta'(0)}{120}+\frac{\zeta'(-2)}{12}+\frac{\zeta'(-4)}{24}{\bf\,,} \quad\cdots \\[2mm] & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \end{align}

Find a formula to calculate either (the first term) or (the limit itself) in each case?

$SO_{n}=\left\{\frac{1}{2}\,,\,\frac{7}{24}\,,\,\frac{1}{9}\,,\,\frac{277}{8640}\,,\,\frac{79}{10800}\,,\,\cdots\right\} \quad\&\quad SE_{n}=\left\{\frac{1}{2}\,,\,\frac{3}{8}\,,\,\frac{11}{72}\,,\,\frac{25}{576}\,,\,\frac{137}{14400}\,,\,\cdots\right\}$

Give $N\ge1\rightarrow$ Find $\small\{\color{blue}{SO_{N}}\,\&\,\color{red}{SE_{N}}\}$ OR $\small\{\color{blue}{\sum\frac{\zeta(2n+1)}{(2n+1)(2n+2)\cdots(2n+N+1)}}\,\&\,\color{red}{\sum\frac{\zeta(2n)}{(2n)(2n+1)\cdots(2n+N)}}\}$

• Pst, the derivative of the Riemann zeta function has a closed form for some of these values. – Simply Beautiful Art Dec 13 '16 at 1:14
• Could you please share a reference. It is trivial to calculate the denominators of the terms involving the derivative; basically it is a factorial sequence with some manipulations. Nevertheless, I fail to calculate the first term. – Hazem Orabi Dec 13 '16 at 1:18
• What do you mean? (maybe refresh the page) I've linked a link that shares how to deal with these derivatives of the zeta. – Simply Beautiful Art Dec 13 '16 at 1:19
• Yes, Thanks for the link. Let us say that the sequences on my question is a bit different. – Hazem Orabi Dec 13 '16 at 1:30
• How do you intend to separate the first term from the others? I do not see how this question can be well defined as of yet. – Simply Beautiful Art Dec 13 '16 at 1:35

The Laurent series expansion of $\Gamma(z)$ around its simple pole at $z=-n$ is

$$\Gamma(z) = \frac{(-1)^n}{n!(z+n)} + \frac{(-1)^n \psi(n+1)}{n!} + O(z+n)$$

or

$$\Gamma(z-n) = \frac{(-1)^n}{n!z} + \frac{(-1)^n \psi(n+1)}{n!} + O(z)$$

as $z \to -n$ (or $z \to 0$) where $\psi(z)$ is the digamma function. (Note that the Taylor series of $\Gamma(z+1)$ around $z=0$ is $\Gamma(z+1) = 1 - \gamma z + O(z^2)$, we'll need this later). We can do the same thing with $\zeta(z)$ at $z=1$ to get

$$\zeta(z) = \frac{1}{z-1} + \gamma + O(z-1)$$

or

$$\zeta(z+1) = \frac{1}{z} + \gamma + O(z)$$

as $z \to 1$ (or $z \to 0$). Now we can use these to evaluate one of your limits:

$$\lim_{x \to 0} \left[ -\frac{1}{2} \Gamma(x-1) - \frac{1}{2!}\Gamma(x+1)\zeta(x+1) \right] = -\frac{1}{2} \lim_{x \to 0} \left[ \Gamma(x-1) + \Gamma(x+1)\zeta(x+1) \right]$$

$$= -\frac{1}{2} \lim_{x \to 0} \left[ \left( -\frac{1}{x} -\psi(2) + O(x) \right) + \left( 1 - \gamma x + O(x^2) \right)\left( \frac{1}{x} + \gamma + O(x) \right) \right]$$

$$= -\frac{1}{2} \lim_{x \to 0} \left[ -\psi(2) + O(x) \right] = \frac{\psi(2)}{2} = \frac{1}{2} - \frac{\gamma}{2}$$

The limits get messier as you go, but all of these Taylor and Laurent series are well known.

• Thanks for the method and the result, it is definitely correct. I edited the question in order to clarify the possibility of finding a general formula to calculate the limit, which is equal to a will define zeta series. – Hazem Orabi Dec 13 '16 at 19:18