<s>[INST] Discussion about $\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)}$ May I have some further discussion on
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)} = \frac{\pi^3}{96} - \frac{\pi}{8}\ln^2\!2 \tag1$$
where $H_{n}$ is harmonic-number and $\psi(x)$ is digamma function.
Formerly, I deduce the result by this elementary approach, also the only solution I have due to the asset for me is quite insufficient.
Start with famous series
$$\frac1{2} \arctan x \ln (1+x^2) = \sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} x^{2n+1}}$$
by which somehow I find (1) actually equal to
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)} = \int_{0}^{1} {\frac{\arctan x \ln(1+x^2)}{x(1+x)} \mathrm{d}x}$$
and this integral is not impossible to solve.
Question 1: Any other possible access to summation as (1)?
For digamma function has strong relation with harmonic-number, here $\psi(n+1)-\psi(n+\tfrac1{2})$ can also be substitution for $H_{2n}-H_{n}$ with some other constants. So what if I want a part of (1) like
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}^{2}}{2n+1}} \quad \text{or} \quad \sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}H_{n}}{2n+1}} \tag2$$
but, of course, I have no idea how to deal with such summation involving higher order of harmonic-number.
Question 2: How to find the closed form for summation in (2)?
 [/INST] Solution to question 1:
Since you managed to write your sum in (1) as $\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx$, so lets evaluate the integral.

From here we have 
$$\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx=\frac{3\pi}{8}\ln^22-\frac{3\pi^3}{96}\tag{1}$$
and from here we have 
$$\int_0^1\frac{\arctan x}x\ln\left(\frac{(1+x^2)^3}{(1+x)^2}\right)dx=0\tag{2}$$
By combining (1) and (2) we get

$$\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^22$$

.

BONUS:
Since 
$$\psi(n+1)-\psi(n+\tfrac12)=H_n-H_{n-\small{\frac12}}=2H_n-2H_{2n}-2\ln2$$
Then the quality in (1) can be written as
$$\small{\sum_{n=0}^\infty\frac{(-1)^nH_{2n}^2}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}-\ln2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}=\frac12\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx}$$
Rearranging
$$\small{\sum_{n=0}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}=\sum_{n=0}^\infty\frac{(-1)^nH_{2n}^2}{2n+1}-\ln2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}-\frac12\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx}$$
Lets evaluate the second sum:
\begin{align}
\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}&=\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}\\
&=\Im\left\{\sum_{n=1}^\infty\frac{(i)^nH_{n}}{n}-\sum_{n=1}^\infty\frac{(i)^n}{n^2}\right\}\\
&=\Im\left\{\frac12\ln^2(1-i)+\operatorname{Li}_2(i)-\operatorname{Li}_2(i)\right\}\\
&=\Im\left\{\frac12\ln^2(1-i)\right\}\\
&=-\frac{\pi}{8}\ln2
\end{align}
So by plugging this result, along with 
$$\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n}^2}{2n+1}=\frac{5}{96}\pi^3+\frac{3\pi}{16}\ln^22+\frac12\ln2\ G+2\Im\left\{\operatorname{Li_3}(1-i)\right\}$$
and 
$$\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^22$$
we finally get

$$\sum_{n=0}^\infty\frac{(-1)^{n}H_{2n}H_n}{2n+1}=\frac{3}{64}\pi^3+\frac{3\pi}{8}\ln^22+\frac12\ln2\ G+2\Im\left\{\operatorname{Li_3}(1-i)\right\}$$

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