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.

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. 
