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 −−−−


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Question Number 165418 by mnjuly1970 last updated on 01/Feb/22

$p r o v e t h a t :$ $1^{} : \underset{n = 1}{\sum^{\infty}} \frac{ζ (2 n) - 1}{1 + n} = \frac{3}{2} - \ln (π)$ $2^{ } : \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} (ζ (n) - 1)}{1 + n} = \frac{3}{2} + \frac{γ}{2} - \frac{\ln (8 π)}{2}$ $3^{ *} : \underset{n = 1}{\sum^{\infty}} \frac{ζ (2 n) - 1}{1 + 2 n} = \frac{3}{2} - \frac{\ln (4 π)}{2}$ $- - - - m . n - - - -$
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