Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 157120 by cortano last updated on 20/Oct/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{(3n+1)(3n+2)}\right)=?$$

Answered by puissant last updated on 20/Oct/21

$$\mathrm{\Omega }=\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{(3n+1)(3n+2)}\right)=\frac{1}{9}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+\frac{1}{3})(n+\frac{2}{3})}$$$$=\frac{1}{9}\left\{\frac{\psi \left(\frac{2}{3}\right)-\psi \left(\frac{1}{3}\right)}{\frac{2}{3}-\frac{1}{3}}\right\}=\frac{1}{3}\{\psi \left(\frac{2}{3}\right)-\psi \left(\frac{1}{3}\right)\}$$$$\bullet \psi (1-z)-\psi \left(z\right)=\pi cotan\left(\pi z\right)forz=\frac{1}{3}$$$$\to \psi \left(\frac{2}{3}\right)=\psi \left(\frac{1}{3}\right)+\pi cotan\left(\frac{\pi }{3}\right)=\psi \left(\frac{1}{3}\right)+\frac{\pi }{\sqrt{3}}..$$$$\Rightarrow \mathrm{\Omega }=\frac{1}{3}\{\psi \left(\frac{1}{3}\right)+\frac{\pi }{\sqrt{3}}-\psi \left(\frac{1}{3}\right)\}=\frac{\pi }{3\sqrt{3}}..$$$$\therefore \because \mathrm{\Omega }=\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{(3n+1)(3n+2)}\right)=\frac{\pi }{3\sqrt{3}}..$$$$$$$$...............\mathcal{L}epuissant...............$$

Commented by Tawa11 last updated on 20/Oct/21

$$\mathrm{great}\mathrm{sir}$$

Answered by cortano last updated on 20/Oct/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{(3n+1)(3n+2)}\right)$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\underset{0}{\stackrel{1}{\int }}(x^{3n}-x^{3n+1})dx$$$$=\underset{0}{\stackrel{1}{\int }}(1-x)\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^{3n}dx$$$$=\underset{0}{\stackrel{1}{\int }}\left(\frac{1-x}{1-x^3}\right)dx=\underset{0}{\stackrel{1}{\int }}\left(\frac{dx}{1+x+x^2}\right)$$$$=\frac{\pi }{3\sqrt{3}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com