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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

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

and

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

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.

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    $\begingroup$ 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 Feb 15 at 9:17
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    $\begingroup$ 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 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 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 at 12:40
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    $\begingroup$ 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 at 22:12
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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 $\sum_{n \ge 1}a_n\frac{H_{2n}}{n}$, $\sum_{n \ge 1}a_n\frac{H_n}{2n+1}$, $\sum_{n \ge 1}a_n\frac{H_{2n}}{2n+1}$.

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

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