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Question Number 148946 by Tawa11 last updated on 01/Aug/21

$$\mathrm{Sum}\mathrm{to}\mathrm{n}\mathrm{term}:\frac{1}{1.2.3}+\frac{1}{4.5.6}+\frac{1}{7.8.9}+...\mathrm{to}\mathrm{n}.$$

Answered by Olaf_Thorendsen last updated on 02/Aug/21

$$\mathrm{S}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(3n+1)(3n+2)(3n+3)}$$$$\mathrm{S}=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{(3n+1)(3n+2)}-\frac{1}{(3n+1)(3n+3)})$$$$\mathrm{S}=\frac{1}{9}\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{(n+\frac{1}{3})(n+\frac{2}{3})}-\frac{1}{(n+\frac{1}{3})(n+1)})$$$$\mathrm{S}=\frac{1}{9}\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{\psi \left(\frac{2}{3}\right)-\psi \left(\frac{1}{3}\right)}{\frac{2}{3}-\frac{1}{3}}-\frac{\psi \left(1\right)-\psi \left(\frac{1}{3}\right)}{1-\frac{1}{3}})$$$$\bullet \psi \left(1\right)=-\gamma$$$$\bullet \psi \left(\frac{1}{3}\right)=-\frac{\pi }{2\sqrt{3}}-\frac{3}{2}\ln 3-\gamma$$$$\bullet \psi (1-z)-\psi \left(z\right)=\pi \cot \left(\pi z\right)$$$$\Rightarrow \psi (1-\frac{1}{3})-\psi \left(\frac{1}{3}\right)=\pi \cot \left(\frac{\pi }{3}\right)$$$$\psi \left(\frac{2}{3}\right)-\psi \left(\frac{1}{3}\right)=\frac{\pi }{\sqrt{3}}$$$$\mathrm{S}=\frac{1}{9}(\frac{\frac{\pi }{\sqrt{3}}}{\frac{1}{3}}-\frac{-\gamma +\frac{\pi }{2\sqrt{3}}+\frac{3}{2}\ln 3+\gamma }{\frac{2}{3}})$$$$\mathrm{S}=\frac{1}{9}(\pi \sqrt{3}-\frac{\pi \sqrt{3}}{4}-\frac{9}{4}\ln 3)$$$$\mathrm{S}=\frac{\pi }{4\sqrt{3}}-\frac{1}{4}\ln 3$$

Commented by Tawa11 last updated on 02/Aug/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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