# The sum of $\sum_{k=0}^{\infty}\frac{\zeta(2k+2)-1}{{2k+1}}$

I don't have an idea how to calculate this sum. I knew that $\sum_{k=0}^{\infty}\zeta(2k+2)-1 = \frac{3}{4}$. I also knew $\sum_{k=1}^{\infty}\frac{\zeta(2k)k-k}{{2k^2+k}}=\frac{1}{2}(3-ln(4\pi))$. Thank you very much for help. The sum to calculate is $$\sum_{k=0}^{\infty}\frac{\zeta(2k+2)-1}{{2k+1}}$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1}:\ {\Large ?}}$.

Note that $\pars{~\mbox{see}\ \color{#000}{\textbf{6.3.14}}\ \mbox{in A & S Table}~}$ $\ds{\left.\vphantom{\Large A} \Psi\pars{1 + z}\right\vert_{\ \verts{z} < 1} = -\gamma + \sum_{n = 2}^{\infty}\pars{-1}^{n}\,\zeta\pars{n}z^{n - 1}}$ where $\ds{\Psi}$ and $\ds{\gamma}$ are the Digamma Function and de Euler Constant, respectively.

Then, with $\ds{\verts{z} < 1}$, \begin{align} \Psi\pars{1 + z} & = -\gamma + \sum_{k = 2}^{\infty}\pars{-1}^{k}\,\zeta\pars{k}z^{k - 1} \\ \Psi\pars{1 - z} & = -\gamma - \sum_{k = 2}^{\infty}\,\zeta\pars{k}z^{k - 1} \end{align} which leads to \begin{align} \Psi\pars{1 + z} - \Psi\pars{1 - z} & = \sum_{k = 2}^{\infty}\bracks{\pars{-1}^{k} + 1}\,\zeta\pars{k}z^{k - 1} = \sum_{k = 0}^{\infty}2\,\zeta\pars{2k + 2}z^{2k + 1} \\[2mm] \mbox{and}\ {\Psi\pars{1 + z} - \Psi\pars{1 - z} \over 2z} - {1 \over 1 - z^{2}} & = \sum_{k = 0}^{\infty}\bracks{\zeta\pars{2k + 2} - 1}z^{2k} \end{align}

Integrating over $\ds{\left[0,1\right)}$:

\begin{align} \sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1} & = \int_{0}^{1}\bracks{% {\Psi\pars{1 + z} - \Psi\pars{1 - z} \over 2z} - {1 \over 1 - z^{2}}}\dd z \\[5mm] & = \int_{0}^{1}\braces{% {\bracks{\Psi\pars{z} + 1/z} - \Psi\pars{1 - z} \over 2z} - {1 \over 1 - z^{2}}}\dd z \\[5mm] & = \int_{0}^{1}\braces{% -\,{\bracks{\Psi\pars{1 - z} - \Psi\pars{z}} - 1/z \over 2z} - {1 \over 1 - z^{2}}}\dd z \\[5mm] & = \int_{0}^{1}\bracks{% -\,{\pi\cot\pars{\pi z} - 1/z \over 2z} - {1 \over 1 - z^{2}}}\dd z \\[5mm] & = {1 \over 2}\int_{0}^{1}\bracks{% {1 - \pi z\cot\pars{\pi z} \over z^{2}} - {2 \over 1 - z^{2}}}\dd z \\[5mm] & = {1 \over 2}\int_{0}^{1}\bracks{% {1 - \pi z\cot\pars{\pi z} \over z^{2}} - {1 \over 1 - z}}\dd z - {1 \over 2}\int_{0}^{1}{\dd z \over 1 + z} \end{align}

$$\bbx{\ds{\sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1} = {1 \over 2}\ \underbrace{\int_{0}^{1}\bracks{% {1 - \pi z\cot\pars{\pi z} \over z^{2}} - {1 \over 1 - z}}\dd z} _{\ds{\mbox{Numerically}\ \approx 2.0463}}\ -\ {1 \over 2}\,\ln\pars{2} \approx 0.6766}}$$

Write $\zeta(2k+2) - 1 = \sum_{n=2}^\infty n^{-2k-2}$ and interchange order of the two sums. I get $$\sum_{n=2}^\infty \frac{\text{arctanh}(1/n)}{n} = \sum_{n=2}^\infty \frac{\ln((n+1)/(n-1))}{2n}$$ However, I still don't have a closed form.

By the integral representation for the $\zeta$ function $$\zeta(m)-1 = \frac{1}{(m-1)!}\int_{0}^{+\infty}\frac{x^{m-1}}{e^x(e^x-1)}\,dx$$ hence $$\begin{eqnarray*} \sum_{k\geq 0}\frac{\zeta(2k+2)-1}{2k+1}&=&\int_{0}^{+\infty}\frac{\sum_{k\geq 0}\frac{x^{2k+1}}{(2k+1)!(2k+1)}}{e^x(e^x-1)}\,dx\\&\stackrel{\text{IBP}}{=}&-\int_{0}^{+\infty}\frac{\sinh x}{x}\left(e^{-x}+\log(1-e^{-x})\right)\,dx\end{eqnarray*}$$ and by exploiting the ordinary generating function for $\{\zeta(2k)\}_{k\geq 1}$ we also have $$\sum_{k\geq 0}\frac{\zeta(2k+2)-1}{2k+1}=\int_{0}^{1}\frac{1-3x^2-\pi x\cot(\pi x)+\pi x^3\cot(\pi x)}{2x^2(1-x^2)}\,dx$$ Still no "nice" closed form, but a simple numerical task, since the last integrand function has an approximately quadratic behaviour on $(0,1)$. Simpson's rule already gives $\text{LHS}\approx\frac{2\pi^2+29}{72}\approx 0.67693$ and by using the composite rule on $5$ points (weights $1-4-2-4-1$) we have $\text{LHS}\approx\frac{29269-6720 \pi +210 \pi ^2}{15120}\approx \color{green}{0.6765}9486.$

• Interesting exercise from a numerical point of view ! – Claude Leibovici Jan 27 '18 at 10:28
• @ClaudeLeibovici: agreed. It is enough to exploit Newton-Cotes formulas together with the computation of $\cot\frac{\pi}{2^n}$. Both can be performed with arbitrary precision, and the outcome can be written in terms of algebraic (constructible) numbers and powers of $\pi$. – Jack D'Aurizio Jan 27 '18 at 15:33
• For sure, I agree with you for the integration. What was more interesting to me was the approximation of the integrand. – Claude Leibovici Jan 27 '18 at 17:33

Starting from Jack D'Aurizio's answer $$\sum_{k\geq 0}\frac{\zeta(2k+2)-1}{2k+1}=\int_{0}^{1}\frac{1-3x^2-\pi x\cot(\pi x)+\pi x^3\cot(\pi x)}{2x^2(1-x^2)}\,dx$$ the integrand can be approximated by a $[2,2]$ Padé approximant built around $x=0$.

Since the first and third derivatives are $0$ for $x=0$, the expression of the Padé approximant of a function $F(x)$ simplifies a lot and write $$F(x)\approx\frac{-12 \left(F(0) F''(0)\right)+ \left(F(0) F''''(0)-6 F''(0)^2\right) x^2} {-12 F''(0)+F''''(0)\, x^2 }$$making $$\int F(x)\,dx=\left(F(0)-\frac{6 F''(0)^2}{F''''(0)}\right)x+\frac{12 \sqrt{3} F''(0)^{5/2} }{F''''(0)^{3/2}}\tanh ^{-1}\left(\frac{\sqrt{F''''(0)} }{2 \sqrt{3F''(0)}}x\right)$$ In the considered problem, we have $$F(0)=\frac{\pi ^2}{6}-1 \qquad F''(0)=\frac{\pi ^4}{45}-2\qquad F''''(0)=\frac{8 \pi ^6}{315}-24$$ making $$\frac{1-3x^2-\pi x\cot(\pi x)+\pi x^3\cot(\pi x)}{2x^2(1-x^2)}\approx \frac{\frac{\pi ^2-6}{6} +\frac{\left(3150 \pi ^2-420 \pi ^4+20 \pi ^6-\pi ^8\right) }{210 \left(\pi ^4-90\right)}x^2 }{1-\frac{2 \left(\pi ^6-945\right) }{21 \left(\pi ^4-90\right)}x^2 }$$ for which the antiderivative is $$\frac{\pi ^2 \left(-3150+420 \pi ^2-20 \pi ^4+\pi ^6\right) x}{20 \left(\pi ^6-945\right)}+\frac{7 \sqrt{{7}} \left(\pi ^4-90\right)^{5/2} \tanh ^{-1}\left(\sqrt{\frac{2 \left(\pi ^6-945\right)}{21 \left(\pi ^4-90\right)}} x\right)}{20 \sqrt 6 \left(\pi ^6-945\right)^{3/2}}$$ Integrating between $0$ and $1$, we get an ugly expression which is $\approx 0.676469$ while the numerical evaluation of Jack D'Aurizio's integral gives $\approx 0.676565$.

Edit

More tedious but still workable, we could build the $[4,2]$ Padé approximant. The result is $\approx 0.676558$.

• @Marianna Kalwat. The maximum difference between the integrand $f$ and its $[2,2]$ Padé approximant $g$ is for $x=1$; this difference is $-\frac{37800-1575 \pi ^4+140 \pi ^6-2 \pi ^8}{420 \pi ^4-40 \pi ^6}\approx 0.000771$. Now $$\int_0^1(f-g)^2\,dx \approx 3.87587\times 10^{-8}$$ – Claude Leibovici Jan 27 '18 at 9:41
• Hello @Claude: since you enjoy playing with that let's notice that Jack D'Aurizio's integral may be rewritten as $\displaystyle\frac 12\int_0^1\frac 1{x^2}-\frac {2}{1-x^2}-\pi\;\frac{\cot(\pi\;x)}x\,dx$. This may help your approximation and mma may have better ideas for the last term. A great year anyway! – Raymond Manzoni Jan 27 '18 at 10:35
• @RaymondManzoni. long time no speak. Not too lat to wish you an happy new year. I shall look at it. Thanks and cheers. – Claude Leibovici Jan 27 '18 at 11:31