Evaluating the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer

The exact value of $\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}$

We have already found that $$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}=\int_0^\infty\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)}$$ if we can evaluate the integral, we are done.
First of all, we need to find the roots $\{l,m,n\}$ of $x^3+3x^2+2x+1$ by Cardano's method. We can obtain: $$l=-\frac{\sqrt[3]{\frac{1}{2} \left(9-\sqrt{69}\right)}}{3^{2/3}}-\sqrt[3]{\frac{2}{3 \left(9-\sqrt{69}\right)}}-1$$ $$m=\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(9-\sqrt{69}\right)}}{2\ 3^{2/3}}+\frac{1-i \sqrt{3}}{2^{2/3} \sqrt[3]{3 \left(9-\sqrt{69}\right)}}-1$$ $$n=\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(9-\sqrt{69}\right)}}{2\ 3^{2/3}}+\frac{1+i \sqrt{3}}{2^{2/3} \sqrt[3]{3 \left(9-\sqrt{69}\right)}}-1$$ Now we need to break down the integrand. Let $a=-l$, $b=-(m+n)$ and $c=mn$ . Then, $x^3+3x^2+2x+1=(x+a)(x^2+bx+c)$ .
By partial fraction,
\begin{align} \frac{2x}{(x+1)(x^3+3x^2+2x+1)} & = \frac{\alpha}{1+x}+\frac{\beta}{a+x}+\frac{\gamma+\delta x}{x^2+bx+c} \\ & = \frac{\alpha}{1+x}+\frac{\beta}{a+x}+\frac{2x\delta+b\delta}{2x^2+2bx+2c}+\frac{-b\delta+2\gamma}{2x^2+2bx+2c} \end{align} where $\alpha=\frac{2}{(1-b+c)(1-a)}$, $\beta=\frac{2a}{(a^2-ab+c)(1-a)}$, $\gamma=\frac{2(1+a-b)c}{(a^2-ab+c)(1-b+c)}$ and $\delta=\frac{2a-2c}{(a^2-ab+c)(1-b+c)}$.
Now we can integrate $\frac{2x}{(x+1)(x^3+3x^2+2x+1)}$ . Since $\int_0^\infty \frac{dx}{x^2+2ax+b} =\frac{1}{\sqrt{b-a^2}}\left(\frac{\pi}{2}-\arctan\left(\frac{a}{\sqrt{b-a^2}}\right)\!\right)$, we can have \begin{align} \sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}} & = \int_0^\infty\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)} \\ & = \int_0^\infty (\frac{\alpha}{1+x}+\frac{\beta}{a+x}+\frac{2x\delta+b\delta}{2x^2+2bx+2c}+\frac{-b\delta+2\gamma}{2x^2+2bx+2c})dx \end{align}

$$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}= -\beta\ln{a}-\frac{1}{2}\delta\ln{c}+(-\arctan(\frac{b}{\sqrt{-b^2+4c}})+\frac{\pi}{2})(\frac{2\gamma-b\delta}{\sqrt{-b^2+4c}})$$

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The exact value of ${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)$

Since $$\frac{(\beta)_k}{(\gamma)_k}=\frac{\Gamma(\gamma)}{\Gamma(\beta)\Gamma(\gamma-\beta)}\int_0^1 t^{\beta-1+k} (1-t)^{\gamma-\beta-1}dt$$ for non-negative integer k, and by the binomial theorem, $$\sum_{k=0}^\infty \frac{(\alpha)_k}{k!}(zt)^k=(1-zt)^{-\alpha}$$ where $0 \le t \le 1$, $-1 \lt z \lt 1$, we have: $${}_2F_1\left(\left.\begin{array}{c} \alpha,\beta\\ \gamma \end{array}\right| z\right)=\frac{\Gamma(\gamma)}{\Gamma(\beta)\Gamma(\gamma-\beta)}\int_0^1 t^{\beta-1}(1-t)^{\gamma-\beta-1}(1-zt)^{-\alpha}dt$$ So, $${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)=\frac{\Gamma(\frac{3}{2})}{\Gamma(\frac{1}{2})}\int_0^1 \frac{dt}{\sqrt{1-t}(1-t/4)}$$ Since $\Gamma(z+1)=z\Gamma(z)$ and the integral can be easily calculated, we finally obtain

$${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)=\frac{2\pi}{3\sqrt{3}}$$

I'm late for this party but the appearance of the plastic constant's minpoly, or $x^3-x-1=0$, got me interested.

The binomial sum can be expressed as a concise finite sum of logarithms. For $k>1$,

$$\sum_{n=1}^\infty\frac1{n\binom{kn}n} =\sum_{n=1}^k \frac{\ln(1-x_n)}{k-(k-1)x_n}=\int_1^\infty\frac1{x(x^k-x+1)}$$

and where the $x_n$, naturally enough, are the roots of $x^k-x+1=0$. Example, for $k=3$, $$A=\sum_{n=1}^\infty\frac1{n\binom{3n}n} = \frac{\ln(1-x_1)}{3-2x_1}+ \frac{\ln(1-x_2)}{3-2x_2}+ \frac{\ln(1-x_3)}{3-2x_3}=0.371216\dots$$ and the $x_n$ are the three roots of $x^3-x+1=0$, one of which is the negated plastic constant $x\approx-1.32472$. It was pointed out that equivalently, $$3A ={_3F_2}\left(\frac32,1,1;\ \frac43,\frac53;\ \frac4{27}\right)$$ Interestingly, the plastic constant also appears in a similar generalized hypergeometric function, $$2B ={_3F_2}\left(\frac12,1,1;\ \frac43,\frac53;\ \frac4{27}\right)$$ discussed by Reshetnikov in this post.

For the most general case,

$\sum\limits_{n=1}^\infty\dfrac{1}{n\binom{kn}{n}}$

$=\sum\limits_{n=1}^\infty\dfrac{\Gamma(n+1)\Gamma((k-1)n+1)}{n\Gamma(kn+1)}$

$=\sum\limits_{n=1}^\infty\dfrac{\Gamma(n)\Gamma((k-1)n+1)}{\Gamma(kn+1)}$

$=\sum\limits_{n=0}^\infty\dfrac{\Gamma(n+1)\Gamma((k-1)n+k)}{\Gamma(kn+k+1)}$

$=~_3\Psi_1\left[\begin{matrix}(1,1)~~(1,1)~~(k,k-1)\\(k+1,k)\end{matrix};1\right]$ (according to http://en.wikipedia.org/wiki/Fox%E2%80%93Wright_function)