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Let the beta and the zeta function be defined as usual:

\begin{align} & \beta(s) & = & \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} & = & 1-\frac{1}{3^s}+\frac{1}{5^s}\dots +\frac{(-1)^n}{(2n+1)^s}+\dots \\ & \zeta(s) & = & \sum_{n=1}^\infty \frac{1}{n^s} &= & 1+\frac{1}{2^s}+\frac{1}{3^s}+\dots +\frac{1}{n^s}+\dots \end{align}

Question

Is there a closed form for $$\sum_{n=1}^\infty \frac{1-\beta(n)}{n}$$

Exposition

I was hunting for some analogues to the zeta function while reading this. In the post we can find

$$ \begin{align} & \sum_{n=2}^\infty \zeta(n)-1 & = & 1 \\ & \sum_{n=2}^\infty \zeta(2n)-1 & = & 3/4 \\ & \sum_{n=2}^\infty \frac{\zeta(n)-1}{n} & = & 1-\gamma \end{align} $$

Where $\gamma$ is the Euler-Mascheroni constant

So I found analogues for the first two but I can't quite see if this is possible for the third. I found using the techniques of the accepted answer in the link above (and it can also be found here):

$$\sum_{n=1}^\infty 1-\beta(n)= \ln(\sqrt2) $$ and $$ \sum_{n=1}^\infty 1-\beta(2n)= \ln(\sqrt2)-\frac{1}{4} $$

Possibly a red herring but hopefully not:

$$\sum_{n=1}^\infty \frac{1-\beta(n)}{n} \approx \ln (L/2)=\ln\bigg( \int_0^1 \frac{dx}{\sqrt{1-x^4}} \bigg) $$ These match up for at least the first 100 decimal places. Where $L=2.62205755429 \dots $ is the Lemniscate constant. If these two are equal my question becomes: What is the connection between the Lemniscate constant and the $\beta$ function?

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  • $\begingroup$ This computation regarding the Lemniscate constant was done by using wolfram alpha: N[1 - DirichletBeta[1] + Sum[(1 - DirichletBeta[n])/n, {n, 2,2000}], 100]. and I compared it to this. $\endgroup$
    – Mason
    Commented Apr 23, 2020 at 1:11
  • $\begingroup$ $$\int_{0}^{\infty}\frac{e^{x}-1}{\left(e^{2x}+1\right)xe^{x}}dx=\ln\bigg( \int_0^1 \frac{dx}{\sqrt{1-x^4}} \bigg) $$ I wonder if this can be shown directly... $\endgroup$
    – Mason
    Commented Apr 6, 2021 at 6:17

1 Answer 1

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First of all : $$ \prod_{p=1}^{n}{\left(1+\frac{1}{4p}\right)}\prod_{p=0}^{n}{\left(1-\frac{1}{4p+3}\right)}=\frac{4\Gamma\left(\frac{7}{4}\right)\Gamma\left(n+\frac{5}{4}\right)\Gamma\left(n+\frac{3}{2}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)\Gamma\left(n+1\right)\Gamma\left(n+\frac{7}{4}\right)}\underset{n\to +\infty}{\longrightarrow}\frac{4\Gamma\left(\frac{7}{4}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)} $$

Let $ n $ be a positive integer, we have the following : \begin{aligned}\sum_{p=1}^{2n+1}{\left(-1\right)^{p}\ln{\left(1+\frac{1}{2p}\right)}}&=\sum_{k=1}^{n}{\ln{\left(1+\frac{1}{4p}\right)}}-\sum_{k=0}^{n}{\ln{\left(1+\frac{1}{4p+2}\right)}}\\&=\ln{\left(\prod_{k=1}^{n}{\left(1+\frac{1}{4p}\right)}\prod_{k=0}^{n}{\left(1-\frac{1}{4p+3}\right)}\right)} \end{aligned}

Thus, $ \sum\limits_{p\geq 0}{\left(-1\right)^{p}\ln{\left(1+\frac{1}{2p}\right)}} $ converges, and its sum values $ \ln{\left(\frac{4\Gamma\left(\frac{7}{4}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)}\right)}\cdot $

Now, since $ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n-1}}{\left(2n+1\right)^{s}}} $ converges absolutly, for any $ s>1 $, we can prove that the family $ \left(\frac{\left(-1\right)^{p-1}}{n\left(2p+1\right)^{n}}\right)_{\left(p,n\right)\in\mathbb{N}^{*}\times\mathbb{N}^{*}\setminus\left\lbrace 1\right\rbrace} $ is summable, and hence, thanks to Fubini's theorem, we can write the following : \begin{aligned}\sum_{n=2}^{+\infty}{\frac{1-\beta\left(n\right)}{n}}=\sum_{n=2}^{+\infty}{\sum_{p=1}^{+\infty}{\frac{\left(-1\right)^{p-1}}{n\left(2p+1\right)^{n}}}}=\sum_{p=1}^{+\infty}{\sum_{n=2}^{+\infty}{\frac{\left(-1\right)^{p-1}}{n\left(2p+1\right)^{n}}}}\end{aligned}

After switching the summations, adding the first term of the sum, recognising logarithm's series expansion, we get that : $$ \sum_{n=1}^{+\infty}{\frac{1-\beta\left(n\right)}{n}}=\sum_{p=1}^{+\infty}{\left(-1\right)^{p-1}\ln{\left(1+\frac{1}{2p}\right)}}=\ln{\left(\frac{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)}{4\Gamma\left(\frac{7}{4}\right)}\right)} $$

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  • $\begingroup$ $$\prod_{k=1}^x \frac{y}{k}+1= \frac{\Gamma(1+x+y)}{\Gamma(1+x)\Gamma(1+y)}$$ and more generally we have this. The product above seems to be indexed in $k$ but it should be $p$. $\endgroup$
    – Mason
    Commented Apr 5, 2021 at 21:29
  • $\begingroup$ This is indeed related to $\int_0^1 \frac{1}{\sqrt{1-x^4}}$. Another wolfram alpha computation. $\endgroup$
    – Mason
    Commented Apr 5, 2021 at 22:00
  • $\begingroup$ @Mason Thanks. In deed the index shoud be $ p $. $\endgroup$
    – CHAMSI
    Commented Apr 6, 2021 at 10:56
  • $\begingroup$ Note that $$ \frac{{3\sqrt \pi }}{4}\frac{{\Gamma \left( {\frac{5}{4}} \right)}}{{\Gamma \left( {\frac{7}{4}} \right)}} = \frac{{\sqrt \pi }}{4}\frac{{\Gamma \left( {\frac{1}{4}} \right)}}{{\Gamma \left( {\frac{3}{4}} \right)}} = \frac{1}{4}\frac{{\Gamma ^2 \left( {\frac{1}{4}} \right)}}{{\sqrt 2 \Gamma \left( {\frac{1}{2}} \right)}} = \frac{1}{{4\sqrt {2\pi } }}\Gamma ^2 \left( {\frac{1}{4}} \right) = \frac{L}{4}. $$ $\endgroup$
    – Gary
    Commented Apr 6, 2021 at 11:05

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