With γ being the Euler Mascheroni constant, this series is well known:1−∑n=2∞ζ(n)−1n=γ (1)The following series involving ζ(2n+1) also seems to converge to the same value, albeit slower:1−∑n=1∞ζ(2n+1)(n+1)(2n+1)=γ (2)Is there a way to derive (2) from (1) ?

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cassidacc4 · Accepted Answer

Note thatζ(2n+1)(n+1)(2n+1)=2ζ(2n+1)−12n+1−ζ(2n+1)−1n+1+22n+1−22n+2Hence∑n=1∞ζ(2n+1)(n+1)(2n+1)=2∑n=1∞ζ(2n+1)−12n+1−∑n=1∞ζ(2n+1)−1n+1+2∑n=3∞(−1)n−1nFinally we apply Question about series involving zeta function? and A Tough Series ∑k=1∞ζ(2k+1)−1k+1=−γ+log⁡(2):∑n=1∞ζ(2n+1)(n+1)(2n+1)=2(1−γ−log⁡(2)2)−(−γ+log⁡(2))+2(log⁡(2)−1+12)=(2−2log⁡(2)−γ)+2log⁡(2)−1=1−γFor a selfcontained proof, just note that2∑n=1∞ζ(2n+1)−12n+1−∑n=1∞ζ(2n+1)−1n+1=∑k=2∞∑n=1∞2(2n+1)k2n+1−∑k=2∞∑n=1∞1(n+1)k2n+1

With \gamma being the Euler Mascheroni constant, this series is

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