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](https://doi.org/10.1016/j.jnt.2013.07.011).



Is it true that $\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k)=\frac23\sum_{k=1}^\infty\frac{\binom{2k}k^2H_{2k}}{(2k+1)16^k}$?
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 $\sum\_{n \ge 1}a\_n\frac{H\_{2n}}{n}$, $\sum\_{n \ge 1}a\_n\frac{H\_n}{2n+1}$, $\sum\_{n \ge 1}a\_n\frac{H\_{2n}}{2n+1}$.


