<s>[INST] Evaluating the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer 
How to evaluate the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer?

For $k=1$, the series does not converge.
When $k=2$, I can prove that:
$$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}=\frac{\pi}{3\sqrt{3}}$$
Usually, this can be proven by differentiating $\sum_{n=1}^\infty \frac{x^{2n}}{n^2 \binom{2n}{n}}=2(\arcsin{\frac{x}{2}})^2$, but I have an alternative proof.

Using the result of:
$$\int_0^\infty \frac{x^ndx}{(x+1)^{y+n+1}}=\frac{1}{y \binom{y+n}{n}} \tag1$$
, which can be easily proved.
I can substitute $y=n$ to obtain
$$\int_0^\infty \frac{x^ndx}{(x+1)^{2n+1}}=\frac{1}{n \binom{2n}{n}}$$
$$\sum_{n=1}^\infty\int_0^\infty \frac{x^ndx}{(x+1)^{2n+1}}=\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}$$
Therefore,
\begin{align}
\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}} & = \int_0^\infty \frac{xdx}{(x+1)(x^2+x+1)} \\
 & = \lim_{L\to \infty}  \frac{1}{2}\ln(x^2+x+1)-\ln(x+1)+\frac{\tan^{-1}(\frac{2x+1}{\sqrt{3}})}{\sqrt{3}}\large{|_0^L} \\
&= \frac{\pi}{3\sqrt{3}}
\end{align}

Now for $k=3$, I tried to substitute $y=2n$ into $(1)$:
$$\int_0^\infty \frac{x^ndx}{(x+1)^{3n+1}}=\frac{1}{2n \binom{3n}{n}}$$
$$\sum_{n=1}^\infty\int_0^\infty \frac{x^ndx}{(x+1)^{3n+1}}=\sum_{n=1}^\infty\frac{1}{2n \binom{3n}{n}}$$
So we can have
$$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}=\int_0^\infty\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)}$$
However, by partial fraction $$\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)}=-\frac{2}{1+x}+\frac{2x^2+4x+2}{x^3+3x^2+2x+1}$$
The left part does not seem to converge.

Feeling frustrated, Wolfram Alpha plays its part. It spits out these results:
$$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}=\frac{1}{3}{}_3F_2\left(\left.\begin{array}{c} 1,1,\frac{3}{2}\\ \frac{4}{3}, \frac{5}{3}  \end{array}\right| \frac{4}{27}\right)$$
$$\sum_{n=1}^\infty\frac{1}{n \binom{4n}{n}}=\frac{1}{4}{}_4F_3\left(\left.\begin{array}{c} 1,1,\frac{4}{3},\frac{5}{3}\\ \frac{5}{4}, \frac{6}{4}, \frac{7}{4}  \end{array}\right| \frac{27}{256}\right)$$
However, I am not very familiar with hypergeometric function.
The pattern suggests that $$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}=\frac{1}{2}{}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2}  \end{array}\right| \frac{1}{4}\right)=\frac{\pi}{3\sqrt{3}}$$
Thus, $${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2}  \end{array}\right| \frac{1}{4}\right)=\frac{2\pi}{3\sqrt{3}}$$

Arriving these results, I have the following questions:  

How can ${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2}  \end{array}\right| \frac{1}{4}\right)$ be expressed into this simple elementary form?
  How can we arrive to the result for $\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}$ given by Wolfram Alpha?
  Ultimately, can we evaluate $\sum_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ for all integers k $\ge$ $2$?

 [/INST] 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)
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