3
$\begingroup$

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 $\small\{\color{blue}{\sum\frac{\zeta(2n+1)}{(2n+1)(2n+2)\cdots(2n+N+1)}}\,\&\,\color{red}{\sum\frac{\zeta(2n)}{(2n)(2n+1)\cdots(2n+N)}}\}$

$\endgroup$
  • 1
    $\begingroup$ Pst, the derivative of the Riemann zeta function has a closed form for some of these values. $\endgroup$ – Simply Beautiful Art Dec 13 '16 at 1:14
  • $\begingroup$ 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. $\endgroup$ – Hazem Orabi Dec 13 '16 at 1:18
  • $\begingroup$ What do you mean? (maybe refresh the page) I've linked a link that shares how to deal with these derivatives of the zeta. $\endgroup$ – Simply Beautiful Art Dec 13 '16 at 1:19
  • $\begingroup$ Yes, Thanks for the link. Let us say that the sequences on my question is a bit different. $\endgroup$ – Hazem Orabi Dec 13 '16 at 1:30
  • $\begingroup$ 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. $\endgroup$ – Simply Beautiful Art Dec 13 '16 at 1:35
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Hazem Orabi Dec 13 '16 at 19:18

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.