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}

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

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)

Again, this is not going to be a full answer. However since my approach is slightly different from what was presented here so far and since my result is similar to the result given by Tito Piezas III I will post my answer. \begin{eqnarray} &&S_k:=\sum\limits_{n=1}^\infty \frac{1}{n \binom{ k n}{n}}=\\ &&\sum\limits_{n=1}^\infty \frac{k n+1}{n} \cdot \int\limits_0^1 t^n \cdot (1-t)^{(k-1) n} dt=\\ &&\int\limits_0^1 \left[ k \frac{t\cdot (1-t)^{k-1}}{1-t \cdot (1-t)^{k-1}} - \log(1-t \cdot (1-t)^{k-1})\right] dt=\\ &&-\int\limits_0^1 t \cdot \frac{ (1-t)^{k-3} (k t-1) \left(t (1-t)^k+k (t-1)+t-1\right)}{\left(t (1-t)^{k-1}-1\right)^2} dt=\\ &&\int\limits_0^1 \frac{(u-1) (k (u-1)+1) u^{k-2} \left(-u^{k-1}+u^k+k+1\right)}{\left(-u^{k-1}+u^k+1\right)^2}du=\\ &&\int\limits_1^\infty \left(\frac{\left(k^2-1\right) (v-1)}{v \left(v^k-v+1\right)}-\frac{k (k+1) (v-1)}{v^2 \left(v^k-v+1\right)}-\frac{k^2 (v-1)^2}{v^2 \left(v^k-v+1\right)^2}+\frac{(k-1) k (v-1)^2}{v \left(v^k-v+1\right)^2}\right) dv=\\ &&\int\limits_1^\infty \frac{(1-v) (k-(k-1) v) \left((k+1) v^k-v+1\right)}{v^2 \left(v^k-v+1\right)^2} dv \end{eqnarray} Let us discuss the steps from the very top to the bottom. In the second line we used the integral representation of the beta function and in the third line we did the sum in question. In the fourth line we integrated by parts to get rid of the logarithm.In the fifth line we substituted for $u:=1-t$ and finally in the sixth and seventh lines we substituted for $v:=1/u$ and simplified the integrand.

Now, oddly enough the equation in the last line is similar to that given by Tito Piezas III. This would suggest that maybe the integrands of both results are the same, which they are not, as I have checked. It would be interesting to simplify my result further and bring it to the form given by Tito Piezas III.

Update: Note that: \begin{equation} \frac{(1+k) v^k-v+1}{v^2(1-v+v^k)^2}= \frac{d}{d v} \left[\frac{1}{v \cdot(1-v+v^k)} \right] + \frac{1}{v (1-v+v^k)^2} \end{equation} Inserting this into the last line above and integrating by parts we get: \begin{eqnarray} S_k&=&\int\limits_1^{\infty} \frac{1}{v \cdot (1-v+v^k)} dv + \int\limits_1^\infty \frac{(k-(k-1)v)}{v \cdot (1-v+v^k)^2} \cdot (-2 v^k-1+v) dv\\ &=&\int\limits_1^{\infty} \frac{1}{v \cdot (1-v+v^k)} dv + \left.\left( \frac{1-v}{1-v+v^k} + \log[\frac{1}{v^k}-\frac{1}{v^{k-1}}+1]\right)\right|_{1}^\infty\\ &=&\int\limits_1^{\infty} \frac{1}{v \cdot (1-v+v^k)} dv+0\\ &=&\int\limits_1^{\infty} \sum\limits_{p=1}^k \frac{1}{v\cdot(v-\xi_p)} \cdot \prod\limits_{q=1,q\neq p}^k\frac{1}{\xi_p-\xi_q)}dv=\\ &=&-\sum\limits_{p=1}^k \frac{\log(1-\xi_p)}{\xi_p} \cdot \prod\limits_{q=1,q\neq p}^k\frac{1}{(\xi_p-\xi_q)} \end{eqnarray} Here $\left\{ \xi_p \right\}_{p=1}^k$ are roots of the polynomial $v^k-v+1$.

As a final comment we note that in general the following holds: \begin{equation} \sum\limits_{n=1}^\infty \frac{x^n}{n} \cdot \frac{1}{\binom{k\cdot n}{n}} =x \cdot \int\limits_1^\infty \frac{1}{v\cdot(v^k+x-x v)} dv \end{equation} where again the integral on the right hand side can be expressed through roots of the polynomial in the denominator.