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Question Number 188381 by mnjuly1970 last updated on 28/Feb/23

$$$$$$\mathrm{Advanced}\mathrm{calculus}$$$$$$$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{series}.$$$$$$$$\mathrm{\Omega }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n\zeta \left(n\right)}{n.2^n}=?$$$$\zeta \left(z\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^z};\mathcal{R}e\left(z\right)>1$$$$$$

Answered by witcher3 last updated on 15/Mar/23

$$\mathrm{n}⩾2$$$$\zeta \left(1\right)\mathrm{diverge}$$

Answered by witcher3 last updated on 18/Mar/23

$$\mathrm{\Omega }=\sum _{\mathrm{n}⩾2}\frac{{(-1)}^{\mathrm{n}}\zeta \left(\mathrm{n}\right)}{\mathrm{n}.2^{\mathrm{n}}}$$$$\mathrm{We}\mathrm{have}\mathrm{\Gamma }\left(\mathrm{n}\right)\zeta \left(\mathrm{n}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx},\mathrm{\forall }\mathrm{n}⩾2$$$$\zeta \left(\mathrm{n}\right)=\frac{1}{(\mathrm{n}-1)!}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{e}}^{\mathrm{x}}-1}\mathbf{dx}$$$$\mathbf{\Omega }=\sum _{\mathrm{n}⩾2}{\int }_0^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}}{2^{\mathrm{n}}.\mathrm{n}!}\mathrm{dx}$$$$\sum _{\mathrm{n}⩾0}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}}{2^{\mathrm{n}}.\mathrm{n}!}\mathrm{cv}\mathrm{Uniformly}\mathrm{to}\exp$$$$\mathrm{this}\mathrm{We}\mathrm{can}\mathrm{exchange}\mathrm{\Sigma }\mathrm{and}\int \mathrm{in}\mathrm{\Omega }$$$$\Rightarrow \mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{x}}\sum _{\mathrm{n}⩾2}\frac{{(-\frac{\mathrm{x}}{2})}^{\mathrm{n}}}{\mathrm{n}!}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\frac{\mathrm{x}}{2}}-1+\frac{\mathrm{x}}{2}}{\mathrm{x}({\mathrm{e}}^{\mathrm{x}}-1)}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}-\frac{{\mathrm{e}}^{-\mathrm{x}}}{\mathrm{x}({\mathrm{e}}^{-\frac{\mathrm{x}}{2}}+1)}+\frac{{\mathrm{e}}^{-\mathrm{x}}}{2(1-{\mathrm{e}}^{-\mathrm{x}})}\mathrm{dx}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{ax}}(-\frac{1}{\mathrm{x}({\mathrm{e}}^{-\frac{\mathrm{x}}{2}}+1)}+\frac{1}{2(1-{\mathrm{e}}^{-\mathrm{x}})})\mathrm{dx}$$$${\mathrm{f}}^{\prime }\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{ax}}(\frac{1}{{\mathrm{e}}^{-\frac{\mathrm{x}}{2}}+1}-\frac{\mathrm{x}}{2(1-{\mathrm{e}}^{-\mathrm{x}})})\mathrm{dx}$$$${\int }_0^{\mathrm{\infty }}-\frac{\mathrm{x}}{2(1-{\mathrm{e}}^{-\mathrm{x}})}{\mathrm{e}}^{-\mathrm{ax}}\mathrm{dx}=-\frac{1}{2}\sum _{\mathrm{k}⩾0}{\int }_0^{\mathrm{\infty }}{\mathrm{xe}}^{-(\mathrm{a}+\mathrm{k})\mathrm{x}}$$$$=-\frac{1}{2}.\mathrm{\Gamma }\left(2\right)\sum _{\mathrm{k}⩾0}\frac{1}{{(\mathrm{k}+\mathrm{a})}^2}=-\frac{1}{2}{\mathrm{\Psi }}^{\left(1\right)}\left(\mathrm{a}\right)$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{ax}}}{{\mathrm{e}}^{-\frac{\mathrm{x}}{2}}+1}=\sum _{\mathrm{k}⩾0}{(-1)}^{\mathrm{k}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{a}+\frac{\mathrm{k}}{2})\mathrm{x}}$$$$=\sum _{\mathrm{k}⩾0}\frac{{(-1)}^{\mathrm{k}}}{(\mathrm{a}+\frac{\mathrm{k}}{2})}=\sum _{\mathrm{k}⩾0}\frac{1}{\mathrm{a}+\mathrm{k}}-\frac{1}{\mathrm{a}+\frac{1}{2}+\mathrm{k}}$$$$=\mathrm{\Psi }(\mathrm{a}+\frac{1}{2})-\mathbf{\Psi }\left(\mathrm{a}\right)$$$${\mathrm{f}}^{\prime }\left(\mathrm{a}\right)=\mathrm{\Psi }(\mathrm{a}+\frac{1}{2})-\mathrm{\Psi }\left(\mathrm{a}\right)-\frac{{\mathrm{\Psi }}^{\left(1\right)}\left(\mathrm{a}\right)}{2}$$$$\mathrm{f}\left(\mathrm{a}\right)=\log \left(\frac{\mathrm{\Gamma }(\mathrm{a}+\frac{1}{2})}{\mathrm{\Gamma }\left(\mathrm{a}\right)}\right)-\frac{1}{2}\mathrm{\Psi }\left(\mathrm{a}\right)+\mathrm{c}$$$$\underset{\mathrm{a}\to \mathrm{\infty }}{\lim }{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{ax}}(-\frac{1}{\mathrm{x}({\mathrm{e}}^{-\frac{\mathrm{x}}{2}}+1)}+\frac{1}{2(1-{\mathrm{e}}^{-\mathrm{x}})})\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}{\underset{\mathrm{a}\to \mathrm{\infty }}{\lim e}}^{-\mathrm{ax}}(-\frac{1}{\mathrm{x}({\mathrm{e}}^{-\frac{\mathrm{x}}{2}}+1)}+\frac{1}{2(1-{\mathrm{e}}^{-\mathrm{x}})})\mathrm{dx}=0$$$$\Rightarrow \underset{\mathrm{a}\to \mathrm{\infty }}{\lim ln}\left(\frac{\mathrm{\Gamma }(\mathrm{a}+\frac{1}{2})}{\mathrm{\Gamma }\left(\mathrm{a}\right)}\right)-\frac{\mathrm{\Psi }\left(\mathrm{a}\right)}{2}+\mathrm{c}=0$$$$\mathrm{\Gamma }\left(\mathrm{a}\right)\sim \sqrt{2\pi }.{\mathrm{a}}^{\mathrm{a}-\frac{1}{2}}{\mathrm{e}}^{-\mathrm{a}}$$$$\mathrm{\Gamma }(\mathrm{a}+\frac{1}{2})\sim \sqrt{2\pi }{\mathrm{a}}^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{a}}$$$$\mathrm{\Psi }\left(\mathrm{a}\right)\sim \ln \left(\mathrm{a}\right)$$$$\underset{\mathrm{a}\to \mathrm{\infty }}{\lim f}\left(\mathrm{a}\right)=\underset{\mathrm{a}\to \mathrm{\infty }}{\lim ln}\left(\sqrt{\mathrm{a}}\right)-\frac{1}{2}\left(\ln \left(\mathrm{a}\right)\right)=0\Rightarrow \mathrm{c}=0$$$$\mathrm{f}\left(\mathrm{a}\right)=\ln \left(\frac{\mathrm{\Gamma }(\mathrm{a}+\frac{1}{2})}{\mathrm{\Gamma }\left(\mathrm{a}\right)}\right)-\frac{\mathrm{\Psi }\left(\mathrm{a}\right)}{2}$$$$\mathrm{\Omega }=\mathrm{f}\left(1\right)=\ln \left(\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }\left(1\right)}\right)-\frac{\mathbf{\Psi }\left(1\right)}{2}=\ln \left(\frac{\sqrt{\pi }}{2}\right)-\frac{\mathrm{\Psi }\left(1\right)}{2}$$$$=\frac{\ln \left(\pi \right)+\gamma }{2}-\ln \left(2\right)=\sum _{\mathrm{n}⩾2}\frac{{(-1)}^{\mathrm{n}}\zeta \left(\mathrm{n}\right)}{2^{\mathrm{n}}.\mathrm{n}}$$$$$$$$$$$$$$$$$$$$$$

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