On the closed form for $\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{5n-1}$ 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 $c_i$? 
 A: 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*}
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\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.

