On Jan. 27, 2012, I conjectured the identity $$\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k)=\frac23\sum_{k=1}^\infty\frac{\binom{2k}k^2H_{2k}}{(2k+1)16^k},\tag{$*$}$$ where $H_n$ denotes the harmonic number $\sum_{k=1}^n\frac1k$. As the two series converge slowly, I lack convincing numerical data to support $(*)$.

Question. Is the identity $(*)$ true? Can one check it further? If it is true, how to prove it?

Your comments are welcome!

Motivation. $(*)$ was motivated by my following conjecture on congruences.

CONJECTURE (Jan 26, 2012). For any prime $p>3$, we have

$$\sum_{k=1}^{(p-1)/2}\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k) \equiv -\frac 73pB_{p-3}\pmod{p^2}\tag{1}$$


$$\sum_{k=1}^{(p-3)/2}\frac{\binom{2k}k^2}{(2k+1)16^k}H_{2k} \equiv -2\left(\frac{-1}p\right)E_{p-3} \pmod p,\tag{2}$$

where $(\frac{\cdot}p)$ is the Legendre symbol, $B_0,B_1,\ldots$ are the Bernoulli numbers and $E_0,E_1,\ldots$ are the Euler numbers.

I noted in 2012 that (2) is equivalent to (1) modulo $p$. See also Conjecture 1.1 of my 2014 JNT paper.

  • 6
    $\begingroup$ Probably this might help $$\sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_n}{16^n}k^{2n}=K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k^2}{16(1-k^2)},\tag{1}$$ $$\sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_{2n}}{16^n}k^{2n}=\frac12K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k}{4(1-k^2)}.\tag{2}$$ $\endgroup$
    – Nemo
    Commented Feb 15, 2019 at 9:17
  • 3
    $\begingroup$ what is the basis for the conjectured equality? a numerical evaluation of the two sides of the equation gives a 1% difference with an error estimate of $10^{-3}$... $\endgroup$ Commented Feb 15, 2019 at 10:58
  • $\begingroup$ If I have done it correctly, you can restate it as $$2+\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(3H_k-2H_{2k})= \sum_{k=1}^\infty\frac{\binom{2k}k^2H_{2k}}{k(2k+1)16^k} $$with better numerical convergence. $\endgroup$
    – Wolfgang
    Commented Feb 15, 2019 at 12:05
  • $\begingroup$ I have heared that Don Zagier has a method to accelerate a slowly convergent series. Is anybody here familiar with the method? $\endgroup$ Commented Feb 15, 2019 at 12:40
  • 1
    $\begingroup$ See also J. Campbell, "New series involving harmonic numbers and squared central binomial coefficients" (2018), hal.archives-ouvertes.fr/hal-01774708, around equation (1.6) for some remarks on the conjecture. $\endgroup$ Commented Feb 15, 2019 at 22:12

1 Answer 1


Let $a_n=\frac{1}{16^n}\binom{2n}n^2$. We have $$ \sum_{n\ge 1} a_n(2H_{2n}-H_n)k^{2n}=-\frac{1}{\pi}K(k)\log(1-k^2). $$ Here $K(x)=\frac{\pi}{2}\sum_{n\ge 0}a_nx^{2n}$ is complete elliptic integral of the first kind. Now devide this identity by $k$ and integrate from $0$ to $1$: \begin{align} \sum_{n\ge 1} \frac{a_n}{2n}(2H_{2n}-H_n)&=-\frac{1}{\pi}\int_0^1K(k)\log(1-k^2)\frac{dk}{k}\\ &=-\frac{1}{2\pi}\int_0^1\frac{\pi}{2}\left(1+\sum_{n\ge 1}a_nx^n\right)\log(1-x)\frac{dx}{x}\\ &=\frac{\pi^2}{24}+\frac14\sum_{n\ge 1}\sum_{m\ge 1}\frac{a_n}{m(n+m)}=\frac{\pi^2}{24}+\frac14\sum_{n\ge 1} \frac{a_n}{n}H_n. \end{align} Thus $$ \sum_{n\ge 1} a_n\frac{4H_{2n}-3H_n}{n}=\frac {\pi^2}{6}.\tag{1} $$ The general series $\, _3F_2(1)$ satisfies $3$-term transformation formula (see Gasper and Rahman, eq. (3.1.3)) \begin{align} \, _3F_2\left({a,b,c\atop d,e};1\right)=\frac{\Gamma (1-a) \Gamma (d) \Gamma (e) \Gamma (c-b)}{\Gamma (c) \Gamma (b-a+1) \Gamma (d-b) \Gamma (e-b)}\, _3F_2\left({b,b-d+1,b-e+1\atop b-c+1,b-a+1};1\right)\\ +\frac{\Gamma (1-a) \Gamma (d) \Gamma (e) \Gamma (b-c)}{\Gamma (b) \Gamma (c-a+1) \Gamma (d-c) \Gamma (e-c)}\, _3F_2\left({c,c-d+1,c-e+1\atop c-a+1,c-b+1};1\right) \end{align} from which one can deduce by setting $e=1$, dividing both sides by $c$ and taking the limit $c\to 0$ $$ \sum_{n\ge 1}\frac{(a)_n(b)_n}{n!(d)_n n}+\psi (1-a)+\psi (b)-\psi (d)-\psi (1)=-\frac{\Gamma (1-a) \Gamma (d)}{b \Gamma (-a+b+1) \Gamma (d-b)}\times\, _3F_2\left({b,b-d+1,b\atop b+1,-a+b+1};1\right),\tag{2} $$ where $\psi$ is digamma function.

Now we apply Newton's method. We have $$ \left\{\frac{d(a)_n}{da}\right\}_{a=1/2}=(1/2)_n\cdot\sum_{k=0}^{n-1} \frac{1}{2k+1}=(1/2)_n(H_{2n}-H_n/2),$$ $$ \left\{\frac{d(a)_n}{da}\right\}_{a=1}=n!\cdot\sum_{k=1}^{n} \frac{1}{k}=n!H_n.$$ First we differentiate (2) wrt to $a$ at $a=b=1/2$, $d=1$ and obtain after simplifications $$ \sum_{n \ge 1}a_n\left(\frac{H_n}{n}+\frac{4H_{2n}-2H_n}{2n+1}\right)=-\frac {\pi^2}{6}+\frac{16 C \log 2}{\pi},\tag{3} $$ where $C$ is catalan's constant.

Similarly by differentiating (2) wrt to $d$ and simplifications we get $$ \sum_{n \ge 1}a_n\left(\frac{2H_{2n}-H_n}{n}+\frac{2H_{n}}{2n+1}\right)=\frac {\pi^2}{2}-\frac{16 C \log 2}{\pi}.\tag{4} $$ Taking the sum of (3) and (4) we get $$ \sum_{n \ge 1}a_n\left(\frac{2H_{2n}}{n}+\frac{4H_{2n}}{2n+1}\right)=\frac {\pi^2}{3}. $$Comparing the last equation and (1) we finally get $$ \sum_{n \ge 1}a_n\left(\frac{2H_{2n}}{n}+\frac{4H_{2n}}{2n+1}\right)=2\cdot \sum_{n\ge 1} a_n\frac{4H_{2n}-3H_n}{n}, $$ which is equivalent to OP's conjecture.

Edit (2019): In the article (see page 20) it was proved that $$ \sum_{n \ge 1}a_n\frac{H_n}{n}=-\frac{5\pi^2}{3}+\frac{64}{\pi}\,\text{Im}\,\text{Li}_3(\tfrac{1+i}{2})+\frac{32}{\pi}C\log 2-2\log^22. $$ Thus equations (1), (3) and (4) allow one to find closed form expressions for the remaining three series $\sum_{n \ge 1}a_n\frac{H_{2n}}{n}$, $\sum_{n \ge 1}a_n\frac{H_n}{2n+1}$, $\sum_{n \ge 1}a_n\frac{H_{2n}}{2n+1}$.

EDIT (2024): The formula $$ \sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_n}{16^n}k^{2n}=K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k^2}{16(1-k^2)},\tag{1} $$ that was provided in the comment to the question, has been proved in the article several years earlier than in the article of Tewodros Amdeberhan, Victor Moll, John Lopez Santander, Ken McLaughlin, Christoph Koutschan. Moreover, proved in a more general form, and in a cleaner way by finding the generating function of the sequence $\frac{(a)_n(1-a)_n}{(n!)^2}H_n$, where $a$ is a continuous parameter.


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