Inspired by Steven Zheng

$S_{0} = \dfrac{1}{2^8}( \dfrac{1}{1^8} + \dfrac{1}{2^8} + \dfrac{1}{3^8} + .........)$

$S_{1} = \dfrac{1}{1^8} + \dfrac{1}{3^8} + \dfrac{1}{5^8} ....$

$S_{0} = \dfrac{1}{2^8}( S_{0} + S_{1})$

$\dfrac{S_{0}}{S_{1}} = \dfrac{1}{2^8}( \dfrac{S_{0}}{S_{1}} + 1)$

$\dfrac{a}{b} = \dfrac{1}{2^8 - 1}$

$\boxed{a + b = 2^8}$

So you got the inspiration from my books. No wonder why I don't remember posting a problem like this on Brilliant.

I actually used the method from my findings $\zeta{(n)} = \frac{{2}^{n}}{{2}^{n}-1} \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { \left( 2k-1 \right) }^{ n } } } .$

$S_{0} = \frac{1}{2^{8}}+\frac{1}{4^{8}}+\frac{1}{6^{8}}+\ldots$

$S_{1} = \frac{1}{1^{8}}+\frac{1}{3^{8}}+\frac{1}{5^{8}}+\ldots$

$2^8S_{0} = \frac{1}{1^{8}}+\frac{1}{2^{8}}+\frac{1}{3^{8}}+\ldots$

$2^8S_{0}-S_{0} = \frac{1}{1^{8}}+\frac{1}{3^{8}}+\frac{1}{5^{8}}+\ldots$

$2^8S_{0}-S_{0} = S_{1}$

$(2^8-1)S_{0} = S_{1}$

$\frac{S_{0}}{S_{1}}=\frac{1}{2^8-1}$

$a=1, b=2^8-1, a+b=2^8=256$

$S_0$ = $\frac{1}{2^8}$ [ $\frac{1}{1^8}$ + $\frac{1}{2^8}$ + $\frac{1}{3^8}$ +...] = $\frac{n}{2^8}$ , where ${n}$ = $\displaystyle \sum_{i=1}^\infty$ $\frac{1}{i^8}$ . Now it is clear to see that ${n}$ = $S_0$ + $S_1$ , so $S_0$ = $\frac{S_0 + S_1}{2^8}$ . Rearranging gives: 255 $S_0$ = $S_1$ , so $\frac{S_0}{S_1}$ = $\frac{1}{255}$ . $\therefore$ ${a}$ + ${b}$ = 256

So+S1 can be expressed as, Sum(n=1...infinity) (1/n^8), while So=Sum(n=1...infinity) [1/(2n)^8]=(1/2^8) Sum(n=1...infinity) (1/n^8). Therefore, (So+S1)/So=2^8=256, So+S1=256So, S1=255So. Therefore, So/S1=1/255, from which a=1 & b=255 so that a+b=256