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 $$\sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_n}{16^n}k^{2n}=K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k^2}{16(1-k^2)},\tag{1}$$ $$\sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_{2n}}{16^n}k^{2n}=\frac12K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k}{4(1-k^2)}.\tag{2}$$ $\endgroup$ – user82588 Feb 15 '19 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 $10^{-3}$... $\endgroup$ – Carlo Beenakker Feb 15 '19 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 Feb 15 '19 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? $\endgroup$ – Zhi-Wei Sun Feb 15 '19 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. $\endgroup$ – Timothy Budd Feb 15 '19 at 22:12

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: In the preprint https://arxiv.org/abs/1806.08411 (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 $\sum_{n \ge 1}a_n\frac{H_{2n}}{n}$, $\sum_{n \ge 1}a_n\frac{H_n}{2n+1}$, $\sum_{n \ge 1}a_n\frac{H_{2n}}{2n+1}$.

| cite | improve this answer | |
  • 1
    $\begingroup$ In (1), shouldn't the first $H_n$ be $H_{2n}$? and similarly in the very last formula? $\endgroup$ – Gerry Myerson May 28 '19 at 4:08
  • 1
    $\begingroup$ @GerryMyerson yes it is a typo. $\endgroup$ – user82588 May 28 '19 at 6:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.