Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 142035 by Dwaipayan Shikari last updated on 25/May/21

$$\frac{1}{2^{1/4}}.\frac{3^{1/9}}{4^{1/16}}.\frac{5^{1/25}}{6^{1/36}}.\frac{7^{1/49}}{8^{1/64}}...=exp(-\frac{{\zeta }^{\prime }\left(2\right)}{2}-\frac{{\pi }^2}{12}\log \left(2\right))$$

Answered by mindispower last updated on 25/May/21

$$=\prod _{k⩾1}\frac{{(1+2k)}^{\frac{1}{{(2k+1)}^2}}}{{\left(2k\right)}^{\frac{1}{{\left(2k\right)}^2}}}=\mathrm{\Psi }$$$$ln\left(\mathrm{\Psi }\right)=\sum _{k⩾1}\frac{ln(2k+1)}{(2k+1)}-\sum _{k⩾1}\frac{ln\left(2k\right)}{4k^2}=A-B$$$$\zeta \left(x\right)=\sum _{n⩾1}\frac{1}{n^x},{\zeta }^{\prime }\left(x\right)=-\sum _{n⩾1}\frac{ln\left(n\right)}{n^x}$$$$B=\sum _{k⩾1}\frac{ln\left(2\right)}{4}.\frac{1}{k^2}+\frac{1}{4}\sum _{k⩾1}\frac{ln\left(k\right)}{k^2}=\frac{ln\left(2\right)}{4}\zeta \left(2\right)-\frac{{\zeta }^{\prime }\left(2\right)}{4}$$$$A=\sum _{k⩾1}(\frac{ln\left(k\right)}{k^2}-\frac{ln\left(2k\right)}{4k^2})=-{\zeta }^{\prime }\left(2\right)-B$$$$ln\left(\mathrm{\Psi }\right)=-\frac{1}{2}{\zeta }^{\prime }\left(2\right)-\frac{ln\left(2\right)}{2}\zeta \left(2\right)=-\frac{{\zeta }^{\prime }\left(2\right)}{2}-\frac{{\pi }^{2}ln\left(2\right)}{12}$$$$\mathrm{\Psi }=exp(\frac{-{\zeta }^{\prime }\left(2\right)}{2}-\frac{{\pi }^{2}ln\left(2\right)}{12})$$$$$$

Commented by Dwaipayan Shikari last updated on 25/May/21

$$Thankssir.Great!$$

Commented by mindispower last updated on 26/May/21

$$pleasursir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com