Wow! (Who's up to the challenge? 11)
Calculus
Level
5
\[\displaystyle\sum _{ n=1 }^{ \infty }{ \frac { \zeta (2n) }{ n(2n+1){ 2 }^{ 2n } } } =A\ln(B\pi )-C\]
where \(A,B,C\) are positive integers, then find \(A+B+C\).
This is a part of Who's up to the challenge?.
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