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 $\ge$ $2$?

  • $\begingroup$ hint: you can find the zeros of the denominator by cardanos method, which enables you to perform a partial fraction decompostion and the result will be logs with ugly roots as arguments (the divergencies will cancel in the end) $\endgroup$ – tired Jul 25 '16 at 16:22
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    $\begingroup$ If you have pushed your computations that far, then you are mature enough to understand that there will be no satisfactory answer to your problem. Expressing a result in terms of hypergeometric functions is just a way of hiding the dirt under the carpet, i.e. you give a name to something that you will still not know. A few years of mathematical practice should be enough to convince any intelligent person that the subset of analytically computable expressions is negligible. Most mathematical expressions (even innocent-looking ones) are, sadly, not computable. $\endgroup$ – Alex M. Jul 25 '16 at 16:51
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    $\begingroup$ @Did: It seems that the OP himself is the first not to understand what I've written in that comment: he has placed a bounty on this question, now... Plus 19 upvotes for the question! I believe that Martians are wondering whether there exist intelligent life on Earth... $\endgroup$ – Alex M. Jul 28 '16 at 21:42
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    $\begingroup$ @AlexM. In your comment, I believe you are saying that most hypergeometric functions cannot be expressed into simplier forms. Although the sum cannot be simplified into exact values, there should be a way for the computer to convert the sum into hypergeometric form. So I am looking for the process of conversation. $\endgroup$ – Mc Cheng Jul 28 '16 at 23:55
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    $\begingroup$ @McCheng This part is standard, computing the ratio of consecutive terms as $$\frac1{(n+1){3n+3\choose n+1}}=\frac1{n{3n\choose n}}\cdot\frac4{27}\frac{n(n+\tfrac12)}{(n+\tfrac23)(n+\tfrac13)}$$ hence, using Pochhammer's symbols, $$\frac1{(n+1){3n+3\choose n+1}}=3\left(\frac4{27}\right)^n\frac1{n!}\frac{(1)_n(1)_n(\tfrac12)_n}{{}{}{}{}{}{}{}(\tfrac53)_n(\tfrac43)_n},$$ from which the ${}_3F_2$ formula should be clear. Likewise, each sum $$\sum_{n=1}^\infty\frac1{n{kn\choose n}}$$ can be rewritten as a value of some ${}_kF_{k-1}$ function. But this is a mere rewriting of the series... $\endgroup$ – Did Jul 29 '16 at 1:17

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

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

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  • $\begingroup$ The expression after the substitution is too messy. I am not able to upload it onto this site. $\endgroup$ – Mc Cheng Aug 1 '16 at 11:41
  • $\begingroup$ To get less daunting formulas, one could introduce the notations $$\omega=\sqrt[3]{\frac{3}{2} \left(9-\sqrt{69}\right)}\qquad j=\frac12(1+i\sqrt3)$$ hence $1+j+j^2=0$ and $j^3=1$, and use them to simplify the formulas, for example, unless I am mistaken, $(l,m,n)$ are $$l=-\frac13\omega-\omega^{-1}-1$$ $$m=\frac13j \omega+j^2\omega^{-1}-1$$ $$n=\frac1{3} j^2\omega+j\omega^{-1}-1$$ and so on... (In particular one sees there might be some sign mistakes in your $l$, or in your $m$ and $n$ since one wants $l+m+n=-3$...) $\endgroup$ – Did Aug 1 '16 at 11:59
  • $\begingroup$ @Did I have checked the roots with Mathematica. Maybe I will follow your suggestion and edit my answer later. $\endgroup$ – Mc Cheng Aug 2 '16 at 1:56
  • $\begingroup$ My bad, $\frac12(1\pm i\sqrt3)$ are not third roots of unity, $\frac12(-1\pm i\sqrt3)$ are (a fact which supports your formulas). Coherent notations would be $$j=\frac12(-1+i\sqrt3)$$ hence $$\frac12(1+i\sqrt3)=-j^2\qquad\frac12(1-i\sqrt3)=-j$$ $\endgroup$ – Did Aug 2 '16 at 7:11

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.

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  • $\begingroup$ very interesting ! thanks for showing $\endgroup$ – G Cab Jan 5 '17 at 0:40
  • $\begingroup$ Hey, I'm late to comment, but I have to say: being late to the party is better than not making it to the party at all! $\endgroup$ – Brevan Ellefsen Jan 17 '17 at 19:39
  • $\begingroup$ @Tito Piezas III: Hi, thanks for this answer. I was wondering whether you could help me transform my answer (see below) to the integral representation that you gave.Thanks in advance. $\endgroup$ – Przemo Sep 11 '17 at 12:35

For the most general case,





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

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  • $\begingroup$ Although it is only a rewriting of the sum, I will still give you the reputation for doing so. $\endgroup$ – Mc Cheng Aug 4 '16 at 0:38

Similar techniques are discussed by Borwein and Girgensohn in http://dx.doi.org/10.1007/s00010-005-2774-x , entitled "Evaluations of binomial series" of 2005.

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  • $\begingroup$ It seems that I have to buy or rent the article... $\endgroup$ – Mc Cheng Jul 26 '16 at 0:04
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    $\begingroup$ @McCheng: Not really, it is also freely available. $\endgroup$ – Alex M. Jul 26 '16 at 13:05

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.

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