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$\dfrac{π}{2}\tan(\dfrac{π}{2}x)= \dfrac{1}{(1-x)}-\dfrac{1}{(1+x)}+\dfrac{1}{(3-x)}-\dfrac{1}{(3+x)}+\dfrac{1}{(5-x)}-\dfrac{1}{(5+x)}+\cdots$ Differentiating respect to $x$ for two times gives $\dfrac{π^3}{8}\sec^2(\frac{π}{2}x)\tan(\frac{π}{2}x)= \dfrac{1}{(1-x)^3}-\dfrac{1}{(1+x)^3}+\dfrac{1}{(3-x)^3}-\dfrac{1}{(3+x)^3}-\cdots$ Take $x=\dfrac{1}{3}$ The series becomes $\dfrac{π^3}{8}\sec^2(\dfrac{π}{6})\tan(\dfrac{π}{6})= \dfrac{3^3}{2^3}-\dfrac{3^3}{4^3}+\dfrac{3^3}{8^3}-\dfrac{3^3}{10^3}+\cdots$ $\dfrac{π^3}{6\sqrt{3}}×\dfrac{8}{27} = 1-\dfrac{1}{2^3}+\dfrac{1}{4^3}-\dfrac{1}{5^3}+\cdots$

$1-\dfrac{1}{2^3}+\dfrac{1}{4^3}-\dfrac{1}{5^3}+\dfrac{1}{7^3}-\dfrac{1}{8^3}+\cdots= \dfrac{4π^3}{81\sqrt{3}}$ Answer is $\color{#20A900}\boxed{a+b=4+3=7}$

For above lemma refer this solution