On the closed form for $\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{5n-1}$

Question

From this post, we find,

$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-1}=\frac{3}{2}+a_1\ln a_2+a_3\sqrt{3}\,\arctan\big(a_4\sqrt{3}\big)$$

$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}=\frac{4}{3}+b_1\ln b_2+b_3\sqrt{4}\,\arctan\big(b_4\sqrt{4}\big)$$

where the $a_i$ and $b_i$ are roots of cubics and quartics, respectively. Would it follow that $p=5$ would have the same form,

$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{5n-1}=\frac{5}{4}+c_1\ln c_2+c_3\sqrt{5}\,\arctan\big(c_4\sqrt{5}\big)$$

and the $c_i$ are now roots of quintics, or is it just hasty generalization? If indeed true, then what are the $c_i$?

Answer 1

I think all the ingredient are already given by previous answers, such as this answer.

Following the same idea, if $p \geq 2$ is an integer and $Z_p = \{ \zeta \in \Bbb{C} : \zeta^p + p - 1 = 0 \}$, then

\begin{align*} \sum_{m=0}^{\infty} \prod_{n=1}^{m} \frac{n}{pn-1} = &= 1 + \int_{0}^{1} \frac{p x^{-1/p}}{(x + p - 1)^2} \, dx \\ &= \frac{p}{p-1} + \frac{1}{p-1} \int_{0}^{1} \frac{pu^{p-2}}{u^p + p - 1} \, du \qquad (x = u^p) \\ &= \frac{p}{p-1} + \frac{1}{p-1} \sum_{\zeta \in Z_p} \frac{1}{\zeta} \int_{0}^{1} \frac{du}{u-\zeta} \\ &= \frac{p}{p-1} + \frac{1}{p-1} \sum_{\zeta \in Z_p} \frac{1}{\zeta} \log\left(1 - \frac{1}{\zeta}\right), \end{align*}

where the standard branch cut $\arg z \in (-\pi, \pi)$ is used to define the complex logarithm $\log z$. Then all you have to do is to simplify the summation for each $p$. It is laborious but not impossible to do, using the fact that

$$ Z_p = \{ (p-1)^{1/p} e^{(2k-1)\pi i/p} : k = 1, \cdots, p \}. $$

Indeed, with $r = (p-1)^{1/p}$ and $\theta_k = (2k-1)\pi/p$ we have

$$\begin{split} \sum_{m=0}^{\infty} \prod_{n=1}^{m} \frac{n}{pn-1} & = \frac{p}{p-1} + \frac{1}{(p-1)r} \sum_{k=1}^{p} \bigg[ \cos \theta_k \log\sqrt{\smash[b]{r^2 - 2r\cos\theta_k + 1}} \\ &\hspace{13em} + \sin \theta_k \arctan \bigg(\frac{\sin\theta_k}{r - \cos\theta_k} \bigg)\bigg] \end{split} \tag{*} $$

Comments.

• Since this will result in roughly $\frac{p}{2}$ logarithmic terms (resp. arctangent terms), I see no reason that they merge into a single logarithm term (resp. single arctangent term) involving only algebraic numbers for any $p$.

• Of course, we expect that the representation $\text{(*)}$ reduces to a single log term and a single arctan term when $p = 2, 3, 4, 6$. This is because both $\{\cos \theta_k\}$ and $\{\sin\theta_k\}$ have rank 1 over $\Bbb{Q}$ for these $p$'s. Indeed, when $p = 6$ we have

\begin{align*} \sum_{m=0}^{\infty} \prod_{n=1}^{m} \frac{n}{6n-1} &= \frac{6}{5} + \frac{\sqrt{3}}{10 \cdot 5^{1/6}} \log\bigg( \frac{5^{1/3} - \sqrt{3} \cdot 5^{1/6} + 1}{5^{1/3} + \sqrt{3} \cdot 5^{1/6} + 1} \bigg) \\ &\hspace{5em} + \frac{1}{5^{7/6}} \bigg( \pi - \arctan \bigg( \frac{10-5\cdot 5^{1/3} + 9 \cdot 5^{2/3}}{13\sqrt{5}} \bigg) \bigg) \\ &\approx 1.2441289574532625062\cdots \end{align*}

Answer 2

I suppose that we could assume that these expressions will have the same form since

$$P_3=\prod_{n=1}^m\frac{n}{3n-1}=3^{-m}\frac{ \Gamma \left(\frac{2}{3}\right) \Gamma (m+1)}{\Gamma \left(m+\frac{2}{3}\right)}$$ $$P_4=\prod_{n=1}^m\frac{n}{4n-1}=4^{-m}\frac{ \Gamma \left(\frac{3}{4}\right) \Gamma (m+1)}{\Gamma \left(m+\frac{3}{4}\right)}$$ $$P_5=\prod_{n=1}^m\frac{n}{5n-1}=5^{-m}\frac{ \Gamma \left(\frac{4}{5}\right) \Gamma (m+1)}{\Gamma \left(m+\frac{4}{5}\right)}$$ $$P_6=\prod_{n=1}^m\frac{n}{6n-1}=6^{-m}\frac{ \Gamma \left(\frac{5}{6}\right) \Gamma (m+1)}{\Gamma \left(m+\frac{5}{6}\right)}$$ where a quite clear patterm appears.

This makes the summations to be

$$S_3=\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-1}=\sum_{m=0}^\infty P_3=\, _2F_1\left(1,1;\frac{2}{3};\frac{1}{3}\right)$$ $$S_4=\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}=\sum_{m=0}^\infty P_4=\, _2F_1\left(1,1;\frac{3}{4};\frac{1}{4}\right)$$ $$S_5=\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{5n-1}=\sum_{m=0}^\infty P_5=\, _2F_1\left(1,1;\frac{4}{5};\frac{1}{5}\right)$$ $$S_6=\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{6n-1}=\sum_{m=0}^\infty P_6=\, _2F_1\left(1,1;\frac{5}{6};\frac{1}{6}\right)$$ where a quite clear pattern appears.

Expanding, we then have (before any simplifications)

$$S_3=\frac{3}{2}+\frac{\log \left(1+\frac{1}{2^{2/3}}-\frac{1}{\sqrt[3]{2}}\right)}{4 \sqrt[3]{2}}-\frac{\log \left(1+\frac{1}{\sqrt[3]{2}}\right)}{2 \sqrt[3]{2}}-\frac{\sqrt{3} \cot ^{-1}\left(\frac{1-2 \sqrt[3]{2}}{\sqrt{3}}\right)}{2 \sqrt[3]{2}}$$

$$S_4=\frac{4}{3}+\frac{\log \left(1+\frac{1}{\sqrt{3}}-\frac{\sqrt{2}}{\sqrt[4]{3}}\right)}{3 \sqrt{2} \sqrt[4]{3}}-\frac{\log \left(1+\frac{1}{\sqrt{3}}+\frac{\sqrt{2}}{\sqrt[4]{3}}\right)}{3 \sqrt{2} \sqrt[4]{3}}+\frac{\sqrt{2} \cot ^{-1}\left(\sqrt{2} \sqrt[4]{3} \left(1-\frac{1}{\sqrt{2} \sqrt[4]{3}}\right)\right)}{3 \sqrt[4]{3}}+\frac{\sqrt{2} \cot ^{-1}\left(\sqrt{2} \sqrt[4]{3} \left(1+\frac{1}{\sqrt{2} \sqrt[4]{3}}\right)\right)}{3 \sqrt[4]{3}}$$

$$S_5=\frac{5}{4}-\frac{\log \left(1+\frac{1}{2^{2/5}}\right)}{4\ 2^{2/5}}-\frac{\left(\sqrt{5}-1\right) \log \left(1+\frac{1}{2^{4/5}}-\frac{1-\sqrt{5}}{2\ 2^{2/5}}\right)}{16\ 2^{2/5}}-\frac{\left(-1-\sqrt{5}\right) \log \left(1+\frac{1}{2^{4/5}}-\frac{1+\sqrt{5}}{2\ 2^{2/5}}\right)}{16\ 2^{2/5}}+\frac{\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}} \cot ^{-1}\left(\frac{2^{2/5} \left(1-\frac{1-\sqrt{5}}{4\ 2^{2/5}}\right)}{\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}}}\right)}{2\ 2^{2/5}}+\frac{\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} \cot ^{-1}\left(\frac{2^{2/5} \left(1-\frac{1+\sqrt{5}}{4\ 2^{2/5}}\right)}{\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}}}\right)}{2\ 2^{2/5}}$$

$$S_6=\frac{6}{5}+\frac{\sqrt{3} \log \left(1+\frac{1}{\sqrt[3]{5}}-\frac{\sqrt{3}}{\sqrt[6]{5}}\right)}{10 \sqrt[6]{5}}-\frac{\sqrt{3} \log \left(1+\frac{1}{\sqrt[3]{5}}+\frac{\sqrt{3}}{\sqrt[6]{5}}\right)}{10 \sqrt[6]{5}}+\frac{2 \cot ^{-1}\left(\sqrt[6]{5}\right)}{5 \sqrt[6]{5}}+\frac{\cot ^{-1}\left(2 \sqrt[6]{5} \left(1-\frac{\sqrt{3}}{2 \sqrt[6]{5}}\right)\right)}{5 \sqrt[6]{5}}+\frac{\cot ^{-1}\left(2 \sqrt[6]{5} \left(1+\frac{\sqrt{3}}{2 \sqrt[6]{5}}\right)\right)}{5 \sqrt[6]{5}}$$ Now, for sure, you can recombine the logarithms and the arccotangents terms to arrive to the kind of expressions you posted for $S_3$ and $S_4$ because the coefficients are the same (same for $S_6$).

For $S_5$ and $S_6$, if feasible for the logarithms, it seems that it could be much more tedious with the arccotangents terms.

In my opinion, the simplest "closed" forms should be the hypergeometric functions.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \end{align} Given the Pochhammer symbol $\displaystyle (a)_m =\frac{\Gamma(a+m)}{\Gamma(a)}$, and where $\Gamma(a)$ is the gamma function. Then,

\begin{align} &\sum_{m = 0}^{\infty}\prod_{n = 1}^{m}{n \over 5n - 1} = \sum_{m = 0}^{\infty}{1 \over 5^{m}}\prod_{n = 1}^{m}{n \over n - 1/5} = \color{blue}{1.32062\dots} \\[5mm] = &\ \sum_{m = 0}^{\infty}{1 \over 5^{m}}\,{m! \over \pars{4/5}_{m}} \qquad\qquad\qquad \\[5mm] = &\ 1+\sum_{m = 1}^{\infty}{m \over 5^{m}}\,{\Gamma\pars{m}\Gamma\pars{4/5} \over \Gamma\pars{4/5 + m}} = 1+\sum_{m = 1}^{\infty}{m \over 5^{m}} \int_{0}^{1}t^{m - 1}\pars{1 - t}^{-1/5}\,\dd t \\[5mm]= &\ 1+\int_{0}^{1}\ \pars{\sum_{m = 1}^{\infty}{m \over 5^{m}}\ t^{m - 1}} \pars{1 - t}^{-1/5}\,\dd t \\[5mm]= &\ 1+\int_{0}^{1}\ \pars{5 \over \pars{5 - t}^{2}} \pars{1 - t}^{-1/5}\,\dd t = \color{blue}{1.32062\dots} \end{align}

Indeed, the final integration is rather cumbersome and it doesn't look close to your guess: It has $\ds{\underline{\mbox{four}}\ \ln}$ with different arguments and $\ds{\underline{\mbox{two}}\ \,\mrm{arccot}}$ with different arguments too.