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Question Number 165418 by mnjuly1970 last updated on 01/Feb/22
       prove  that :       1^∗ : Σ_(n=1) ^∞ (( ζ (2n )−1)/( 1+ n)) = (3/(2 ))  − ln (π )     2^( ∗∗) :  Σ_(n=2) ^∞ (( (−1)^( n) (  ζ (n )−1 ))/(1 + n))=(3/2) +(γ/2) −((ln(8π))/2)     3^( ∗∗)  :  Σ_(n=1) ^∞ (( ζ (2n )−1)/(1+ 2n)) = (3/2) −((ln(4π))/2)                −−−− m.n −−−−
$$ \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$ \\ $$$$\:\:\:\mathrm{1}^{\ast} :\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\zeta\:\left(\mathrm{2}{n}\:\right)−\mathrm{1}}{\:\mathrm{1}+\:{n}}\:=\:\frac{\mathrm{3}}{\mathrm{2}\:}\:\:−\:\mathrm{ln}\:\left(\pi\:\right) \\ $$$$\:\:\:\mathrm{2}^{\:\ast\ast} :\:\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\:{n}} \left(\:\:\zeta\:\left({n}\:\right)−\mathrm{1}\:\right)}{\mathrm{1}\:+\:{n}}=\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\gamma}{\mathrm{2}}\:−\frac{\mathrm{ln}\left(\mathrm{8}\pi\right)}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{3}^{\:\ast\ast} \::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\zeta\:\left(\mathrm{2}{n}\:\right)−\mathrm{1}}{\mathrm{1}+\:\mathrm{2}{n}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\mathrm{ln}\left(\mathrm{4}\pi\right)}{\mathrm{2}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−\:{m}.{n}\:−−−− \\ $$$$ \\ $$

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