Question-179918

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

Answered by mindispower last updated on 05/Nov/22

ζ(t+=−γ−Σζ(n+1)(−t)^n Ψ′(z+1)=−Σ((nζ(n+1)(−1)^n t^(n−1) )/1) zΨ′′(z+1)=−Σn(−1)^n t^n ζ(n+1) Ψ′(z+1)+zΨ′′(z+1)=Σn^2 (−1)^(n−1) z^(n−1) ⇒Ψ′((1/2))−((Ψ′((1/2_ )))/2)=Σn^2 ((ζ(n+1))/2^(n−1) ) =(1/4)Ψ′((1/2))−((Ψ′′((1/2)))/8)=S Ψ′((1/2))=(1/((n+z)^2 ))=(4/((2n+1)^2 ))=4(ζ(2)−((ζ(2))/4))=3ζ(2) ((−16)/((2n+1)^2 ))=−16((7/8)ζ(3))=−14ζ(3) (3/4)ζ(2)+(7/4)ζ(3)=S

$$\zeta\left({t}+=−\gamma−\Sigma\zeta\left({n}+\mathrm{1}\right)\left(−{t}\right)^{{n}} \right. \\ $$$$\Psi'\left({z}+\mathrm{1}\right)=−\Sigma\frac{{n}\zeta\left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} {t}^{\boldsymbol{{n}}−\mathrm{1}} }{\mathrm{1}} \\ $$$${z}\Psi''\left({z}+\mathrm{1}\right)=−\Sigma{n}\left(−\mathrm{1}\right)^{{n}} {t}^{{n}} \zeta\left({n}+\mathrm{1}\right) \\ $$$$\Psi'\left({z}+\mathrm{1}\right)+{z}\Psi''\left({z}+\mathrm{1}\right)=\Sigma{n}^{\mathrm{2}} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {z}^{{n}−\mathrm{1}} \\ $$$$\Rightarrow\Psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\Psi'\left(\frac{\mathrm{1}}{\mathrm{2}_{} }\right)}{\mathrm{2}}=\Sigma{n}^{\mathrm{2}} \frac{\zeta\left({n}+\mathrm{1}\right)}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\Psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\Psi''\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{8}}={S} \\ $$$$\Psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\left({n}+{z}\right)^{\mathrm{2}} }=\frac{\mathrm{4}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{4}\left(\zeta\left(\mathrm{2}\right)−\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{4}}\right)=\mathrm{3}\zeta\left(\mathrm{2}\right) \\ $$$$\frac{−\mathrm{16}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=−\mathrm{16}\left(\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)\right)=−\mathrm{14}\zeta\left(\mathrm{3}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)+\frac{\mathrm{7}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)={S} \\ $$$$ \\ $$

Commented by mindispower last updated on 05/Nov/22

$${withe}\:{pleazud}\:{god}\:{bless}\:{You} \\ $$

Commented by mnjuly1970 last updated on 05/Nov/22

$${thanks}\:{alot}\:\:{sir} \\ $$

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