<s>[INST] Evaluations of three series involving binomial coefficients Question. How to prove the following three identities?
\begin{align}\sum_{k=1}^\infty\frac1{k(-2)^k\binom{2k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right)=\frac{\log^22}3-\frac{\pi^2}{36},\tag{1}
\end{align}
\begin{align}\sum_{k=1}^\infty\frac1{k2^k\binom{3k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right)
=\frac{3}{10}\log^22+\frac{\pi}{20}\log2-\frac{\pi^2}{60},\tag{2}
\end{align}\begin{align}\sum_{k=1}^\infty\frac1{k^22^k\binom{3k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right)
=-\frac{\pi G}2+\frac{33}{32}\zeta(3)+\frac{\pi^2}{24}\log2,\tag{3}
\end{align}
where $G$ denotes the Catalan constant $\sum_{k=0}^\infty(-1)^k/{(2k+1)^2}$.
Remark. Motivated by my study of congruences, in 2014 I tried to evaluate the three series in $(1)$-$(3)$, and this led me to discover  $(1)$-$(3)$ which can be easily checked numerically via Mathematica. But I'm unable to prove the above three identities. Also, Mathematica could not evaluate the three series. For more backgrounds of this topic, you may visit http://maths.nju.edu.cn/~zwsun/165s.pdf.
Your comments are welcome!
 [/INST] Here is a proof of the first identity. The others can probably be done similarly.
We have
$$\frac1{k\binom{2k}{k}}=\frac12 B(k,k)=\frac12 \int_0^1 t^{k-1}(1-t)^{k-1}\,dt,$$
$$\frac{1}{k+1}+\cdots+\frac{1}{2k} = \int_0^1 \frac{1-x^{2k}}{1+x}\, dx$$
and
$$\sum_{k=1}^\infty \frac{1}{(-2)^k}t^{k-1}(1-t)^{k-1}(1-x^{2k}) = -\frac{1}{2+t(1-t)}+\frac{x^2}{2+t(1-t)x^2}.$$
Combining all these together, we get
$$\sum_{k=1}^\infty\frac1{k(-2)^k\binom{2k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right) = I_1 + I_2,$$
where
$$I_1 := -\int_0^1\frac{dx}{2(1+x)}\int_0^1 dt\,\frac{1}{2+t(1-t)} = -\frac13 \log(2)^2,$$
$$I_2 := \int_0^1 dx \frac{x^2}{2(1+x)} \int_0^1 dt\, \frac{1}{2+t(1-t)x^2} = \frac23\log(2)^2 - \frac{\pi^2}{36}.$$
So,
$$I_1 + I_2 = \frac13\log(2)^2 - \frac{\pi^2}{36}.$$
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