Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}$ how to prove that 

$$\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}\ ?$$

where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.

This problem is proposed by a friend on a Facebook group and I managed to prove the equality using only integration but can we prove it using series manipulations? 
Here is my work,
In page $105$ of this paper we have
$$\overline{H}_n-\ln2=(-1)^{n-1}\int_0^1\frac{x^n}{1+x}dx$$
Therefore  
$$(\overline{H}_n-\ln2)^2=\int_0^1\int_0^1\frac{(xy)^n}{(1+x)(1+y)}dxdy$$
$$\Longrightarrow \sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)}\sum_{n=0}^\infty(-xy)^n$$
$$=\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)(1+xy)}=\int_0^1\frac{1}{1+x}\left(\int_0^1\frac{dy}{(1+y)(1+xy)}\right)dx$$
$$=\int_0^1\frac{1}{1+x}\left(-\frac{\ln\left(\frac{1+x}{2}\right)}{1-x}\right)dx=-\int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx,\quad x=\frac{1-u}{1+u}$$
$$=\frac12\int_0^1\frac{\ln(1+u)}{u}du=\frac12(-\operatorname{Li}_2(-1))=\frac12(\frac12\zeta(2))=\frac{\pi^2}{24}$$

Regarding series manipulation, we know that 
$$\overline{H}_{2n}=H_{2n}-H_n$$ but we need $\overline{H}_n$ , so is this identity helpful? any other ideas? 
thank you
 A: The development of the solution using only series manipulations
I asked Cornel for a way and here you have the large steps to go for perfectly obtaining what you need. You need a mix of results and I'll provide with references for all you need. 
First step

Start with Abel's summation $a_k=(-1)^k$, $b_k=(\overline{H}_k-\log(2))^2$ to show that 
  $$ \sum_{n=0}^\infty(-1)^n(\overline{H}_n-\log(2))^2=2 \sum _{n=0}^{\infty } \frac{H_n-H_{2 n}+\log (2)}{2 n+1}-\frac{\pi ^2}{8}.$$

Second step

This is a non-obvious step. Build the series below
  $$\sum _{n=0}^{\infty } \frac{H_n-H_{2 n}+\log (2)}{(2 n+1) (2 n+2)}=\frac{1}{12} \left(\pi ^2+6 \log ^2(2)-12 \log (2)\right).$$
  Apply Abel's summation with $a_k=1/((2k+1)(2k+2))$ and $b_k=H_k-H_{2 k}+\log (2)$ to almost magically get the same sum in the right-hand side, but with an opposite sign together with elementary series to calculate directly.

Third Step

Show that 
  $$\sum_{n=1}^{\infty} \frac{1}{n}(H_{2n}-H_n-\log(2))=\log^2(2)-\frac{\pi^2}{12}.$$
  The strategy of approaching the series is presented on page $250$ in the book (Almost) Impossible Integrals, Sums, and Series. The only thing you need differently here is the approach of $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_n}{n}$, which you want elementary by series manipulations. Therefore, see the next step.

Fourth step

Show that 
  $$ \sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_n}{n} = \frac{\pi^2}{12} - \frac{1}{2} \log^2(2),$$
  which is proved completely elementarily by series manipulations in https://math.stackexchange.com/q/499689.

Putting all together leads to the desired result, that is the evaluation of that series by series manipulations. 
Q.E.D.
A nice note: If we are allowed to use integrals, then we may also use a different approach to calculate the series from the Third step, by exploiting the series

$$\sum_{n=1}^{\infty} \left(2 H_{2n}-2 H_{n}+\frac{1}{2n}-2 \log(2)\right)\frac{\sin^2(2n x)}{n}=\log(\sin(x))\log(\cos(x)), \ 0<x<\frac{\pi}{2},$$

that appears on page $248$ of the previously mentioned book. Then in the extraction process of the desired value we also need to prove that 

$$\int_0^{\pi/2} \log(\sin(x))\log(\cos(x))\textrm{d}x =\frac{\pi}{2}\log ^2(2)-\frac{\pi ^3}{48}.$$

At this point I recall that Paul Nahin presents in his book Inside Interesting Integrals a nice generalization for 

$$\int_0^{\pi/2} \log(a\sin(x))\log(a\cos(x))\textrm{d}x, \ a>0,$$

which one may find on page $236$. Also, there could be a temptation to immediately treat the integral with $a=1$ as a particular case of the Beta function in the trigonometric form. According to Inside Interesting Integrals, the last integral together with the forms $\int_0^{\pi/2} \log^2(a\sin(x))\textrm{d}x, \ a>0,$ and $\int_0^{\pi/2} \log^2(a\cos(x))\textrm{d}x, \ a>0,$ were evaluated due to the English mathematician Joseph Wolstenholme.
