<s>[INST] Evaluate $\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$ By accident, I find this summation when I pursue the particular value of $-\operatorname{Li_2}(\tfrac1{2})$, which equals to integral $\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x}$.
Notice this observation
$$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x} = \int_{0}^{1} {\frac{\ln(1-x^{2})}{1+x} \mathrm{d}x} - \frac{(\ln2)^{2}}{2}$$
And using the Taylor series of $\ln(1-x^{2})$, I found this summation $\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$, where $H_{n}$ is the harmonic-numbers.
If the value of $\operatorname{Li_2}(\tfrac1{2})=\tfrac1{2}(\zeta(2)-(\ln2)^{2})$ is given, the result can be easily deduced, which is

$$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)} = -\frac{\zeta(2)}{2}+(\ln2)^{2}$$

For the original goal is to calculate $\operatorname{Li_2}(\tfrac1{2})$, I expect some other approaches to the summation without using the value of $\operatorname{Li_2}(\tfrac1{2})$. I already knew some famous problem like Euler's Sum, which holds very similar form to this summation, but still in trouble finding the appropriate path.
 [/INST] Here is an approach that avoids knowing the value of $\operatorname{Li}_2 (\frac{1}{2})$. 
Let
$$S = \sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ).$$
Observing that
$$\int_0^1 \frac{x^{2n}}{1 + x} \, dx = H_n - H_{2n} + \ln 2,$$
your sum can be rewritten as
\begin{align}
S &= -\int_0^1 \frac{1}{1 + x} \sum_{n = 1}^\infty \frac{x^{2n}}{n} \, dx\\
&= \int_0^1 \frac{\ln (1 - x^2)}{1 + x} \, dx\\
&= \int_0^1 \frac{\ln (1 + x)}{1 + x} \, dx + \int_0^1 \frac{\ln (1 - x)}{1 + x} \, dx\\
&= I + J.
\end{align}
For the first of the integrals $I$, one has
$$I = \frac{1}{2} \ln^2 2.$$
Now consider $J - I$. Then
$$J - I = \int_0^1 \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{1 + x}.$$
Employing a self-similar substitution of $t = (1-x)/(1+x)$ leads to
\begin{align}
J - I &= \int_0^1 \frac{\ln t}{1 + t} \, dt\\
&= \sum_{n = 0}^\infty (-1)^n \int_0^1 t^n \ln t \, dt\\
&= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\int_0^1 t^{n + s} \, dt \right ]_{s = 0}\\
&= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\
&= -\underbrace{\sum_{n = 0}^\infty \frac{(-1)^n}{(n + 1)^2}}_{n \, \mapsto \, n - 1}\\
&= -\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^2}\\
&= -\sum_{n = 1}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\
&= -\frac{1}{2} \zeta (2).
\end{align} 
Thus
$$J = I - \frac{1}{2} \zeta (2) = \frac{1}{2} \ln^2 2 - \frac{1}{2} \zeta (2).$$
Since $S = I + J$, we immediately see that
$$\sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ) = \ln^2 2 - \frac{1}{2} \zeta (2),$$
as desired. 
</s>