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 $\ge$ $2$?
Best Answer
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}
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