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$$\begin{array}{rl}& 2\sum _{0<a}\frac{2^a}{a^2(\genfrac{}{}{0ex}{}{2a}{a})}\\ & =\sum _{0<a<b}\frac{2^a}{ab}\frac{a!b!}{(a+b)!}+\sum _{0<a}\frac{2^a}{a^2(\genfrac{}{}{0ex}{}{2a}{a})}\\ & =\sum _{0<a\le b}\frac{2^a}{ab}\frac{a!b!}{(a+b)!}\\ & =\sum _{0<b}\frac{b!}{b}\sum _{a=1}^b\frac{2^{a}a!}{a(a+b)!}\\ & =\sum _{0<b}\frac{b!}{b}\frac{1}{2^{b}b!}\sum _{a=1}^b\frac{2^a}{a}\\ & =\sum _{0<a\le b}\frac{2^a}{ab{2}^b}\\ & =\sum _{0<a}\frac{2^a}{a}\sum _{a<b}\frac{1}{b{2}^b}+\zeta (2)\\ & =\sum _{0<a}\frac{2^a}{a}(-\frac{a!}{2^a}\sum _{0<b}\frac{(-1{)}^{b}b!}{b(a+b)!})+\zeta (2)\\ & =-\sum _{0<b}\frac{(-1{)}^{b}b!}{b}\sum _{0<a}\frac{a!}{a(a+b)!}+\zeta (2)\\ & =\zeta (2)-\zeta (\overline{2})\\ & =\frac{{\pi }^2}{4}\end{array}$$