On Jan. 27, 2012, I conjectured the identity $$\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k)=\frac23\sum_{k=0}^\infty\frac{\binom{2k}k^2H_{2k}}{(2k+1)16^k},\tag{$*$}$$$$\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k)=\frac23\sum_{k=1}^\infty\frac{\binom{2k}k^2H_{2k}}{(2k+1)16^k},\tag{$*$}$$ 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
$$\sum_{k=0}^{(p-3)/2}\frac{\binom{2k}k^2}{(2k+1)16^k}H_{2k} \equiv -2\left(\frac{-1}p\right)E_{p-3} \pmod p,\tag{2}$$$$\sum_{k=1}^{(p-3)/2}\frac{\binom{2k}k^2}{(2k+1)16^k}H_{2k} \equiv -2\left(\frac{-1}p\right)E_{p-3} \pmod p,\tag{2}$$
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.