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Question Number 179918 by mnjuly1970 last updated on 04/Nov/22

Answered by mindispower last updated on 05/Nov/22

$$\zeta (t+=-\gamma -\mathrm{\Sigma }\zeta (n+1){(-t)}^n$$$${\mathrm{\Psi }}^{\prime }(z+1)=-\mathrm{\Sigma }\frac{n\zeta (n+1){(-1)}^n{t}^{\mathit{n}-1}}{1}$$$$z{\mathrm{\Psi }}^{″}(z+1)=-\mathrm{\Sigma }n{(-1)}^n{t}^n\zeta (n+1)$$$${\mathrm{\Psi }}^{\prime }(z+1)+z{\mathrm{\Psi }}^{″}(z+1)=\mathrm{\Sigma }{n}^2{(-1)}^{n-1}{z}^{n-1}$$$$\Rightarrow {\mathrm{\Psi }}^{\prime }\left(\frac{1}{2}\right)-\frac{{\mathrm{\Psi }}^{\prime }\left(\frac{1}{2_{}}\right)}{2}=\mathrm{\Sigma }{n}^2\frac{\zeta (n+1)}{2^{n-1}}$$$$=\frac{1}{4}{\mathrm{\Psi }}^{\prime }\left(\frac{1}{2}\right)-\frac{{\mathrm{\Psi }}^{″}\left(\frac{1}{2}\right)}{8}=S$$$${\mathrm{\Psi }}^{\prime }\left(\frac{1}{2}\right)=\frac{1}{{(n+z)}^2}=\frac{4}{{(2n+1)}^2}=4(\zeta \left(2\right)-\frac{\zeta \left(2\right)}{4})=3\zeta \left(2\right)$$$$\frac{-16}{{(2n+1)}^2}=-16\left(\frac{7}{8}\zeta \left(3\right)\right)=-14\zeta \left(3\right)$$$$\frac{3}{4}\zeta \left(2\right)+\frac{7}{4}\zeta \left(3\right)=S$$$$$$

Commented by mindispower last updated on 05/Nov/22

$$withepleazudgodblessYou$$

Commented by mnjuly1970 last updated on 05/Nov/22

$$thanksalotsir$$

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