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∑n=1∞(ζ(2n)n−12n−12n+1)\large \sum_{n=1}^{\infty} \left(\dfrac{\zeta(2n)}{n} - \dfrac{1}{2n} - \dfrac{1}{2n+1} \right) n=1∑∞(nζ(2n)−2n1−2n+11)
Find the value of the closed form of the above series.
Give your answer to 1 decimal place.
Notation: ζ(⋅)\zeta(\cdot) ζ(⋅) denotes the Riemann zeta function.
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