TITLE: Evaluate the series $\sum_{k\geq 1} \frac{1}{2^k k^2}$
QUESTION [9 upvotes]: How to prove that  
$$\sum_{k\geq 1} \frac{1}{2^k k^2}=\frac{\pi^2}{12}-\frac{1}{2}\log(2)^2$$
without using the well-known $\operatorname{Li}_2\left( \frac{1}{2} \right)$ ?
Edited : Thanks for L.F , but I should have made it clear that I want an answer using only series manipulations .

REPLY [6 votes]: Apply Euler Series Transformation to the series
$$
\frac{\pi^2}{12}=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}\tag{1}
$$
which yields
$$
\begin{align}
\frac{\pi^2}{12}
&=\sum_{n=0}^\infty2^{-n-1}\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{(k+1)^2}\tag{2}\\
&=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{k=0}^n\binom{n+1}{k+1}\frac{(-1)^k}{k+1}\tag{3}\\
&=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{k=0}^n\sum_{j=0}^n\binom{j}{k}\frac{(-1)^k}{k+1}\tag{4}\\
&=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{j=0}^n\sum_{k=0}^n\binom{j+1}{k+1}\frac{(-1)^k}{j+1}\tag{5}\\
&=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{j=0}^n\frac1{j+1}\tag{6}\\
&=\sum_{n=1}^\infty2^{-n}\frac1{n}\sum_{j=1}^n\frac1j\tag{7}\\
&=\sum_{n=1}^\infty\frac{H_n}{n2^n}\tag{8}
\end{align}
$$
Use the series for $\log(2)=-\log\left(1-\frac12\right)$
$$
\log(2)=\sum_{m=1}^\infty\frac1{k2^k}\tag{9}
$$
and square to get
$$
\begin{align}
\log(2)^2
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{mn2^{m+n}}\tag{10}\\
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(m+n)2^{m+n}}+\frac1{n(m+n)2^{m+n}}\tag{11}\\
&=2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(m+n)2^{m+n}}\tag{12}\\
&=2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac1{mn2^n}\tag{13}\\
&=2\sum_{m=1}^\infty\sum_{n=m}^\infty\frac1{mn2^n}-2\sum_{m=1}^\infty\frac1{m^22^m}\tag{14}\\
&=2\sum_{n=1}^\infty\sum_{m=1}^n\frac1{mn2^n}-2\sum_{m=1}^\infty\frac1{m^22^m}\tag{15}\\
&=2\sum_{n=1}^\infty\frac{H_n}{n2^n}-2\sum_{m=1}^\infty\frac1{m^22^m}\tag{16}\\
&=\frac{\pi^2}{6}-2\sum_{m=1}^\infty\frac1{m^22^m}\tag{17}
\end{align}
$$
Therefore,
$$
\sum_{m=1}^\infty\frac1{m^22^m}=\frac{\pi^2}{12}-\frac12\log(2)^2\tag{18}
$$

Explanation
$\hphantom{0}(2)$ Euler Series Transformation
$\hphantom{0}(3)$ $\binom{n+1}{k+1}=\frac{n+1}{k+1}\binom{n}{k}$
$\hphantom{0}(4)$ $\binom{n+1}{k+1}=\sum_{j=0}^n\binom{j}{k}$
$\hphantom{0}(5)$ $\binom{j+1}{k+1}=\frac{j+1}{k+1}\binom{j}{k}$
$\hphantom{0}(6)$ $\sum_{k=0}^j(-1)^k\binom{j+1}{k+1}=1$
$\hphantom{0}(7)$ $j\mapsto j-1$ and $n\mapsto n-1$
$\hphantom{0}(8)$ collect $H_n$
$(10)$ multiply the series
$(11)$ $\frac1{mn}=\frac1{m(m+n)}+\frac1{n(m+n)}$
$(12)$ use symmetry
$(13)$ $n\mapsto n-m$
$(14)$ add and subtract the terms for $m=n$
$(15)$ change order of summation
$(16)$ collect $H_n$
$(17)$ use $(8)$