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Question Number 121928 by Dwaipayan Shikari last updated on 12/Nov/20

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{12n+1}+\frac{1}{12n+5}-\frac{1}{12n+7}-\frac{1}{12n+11})$$$$$$$$Problemsource:Brilliant.Org$$

Commented by Dwaipayan Shikari last updated on 12/Nov/20

https://brilliant.org/problems/not-quite-a-telescoping-sum-2 https://brilliant.org/problems/integral-out-of-the-box

Commented by Dwaipayan Shikari last updated on 12/Nov/20

$$\frac{1}{12}\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{n+\frac{1}{12}}-\frac{1}{n+\frac{11}{12}}+\frac{1}{n+\frac{5}{12}}-\frac{1}{n+\frac{7}{12}})$$$$=\frac{1}{12}\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{n+\frac{1}{12}}-\frac{1}{n+1-\frac{1}{12}})+\frac{1}{12}\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{n+\frac{5}{12}}-\frac{1}{n+1-\frac{7}{12}})$$$$=\frac{1}{12}(\pi cot\left(\frac{\pi }{12}\right)+\pi cot\left(\frac{5\pi }{12}\right))$$$$=\frac{\pi }{12}(2-\sqrt{3}+2+\sqrt{3})$$$$=\frac{\pi }{3}$$

Answered by mnjuly1970 last updated on 12/Nov/20

$$solution:$$$$S=\sum _{n⩾0}(\frac{1}{12n+1}-\frac{1}{12n+11})+\sum _{n⩾0}(\frac{1}{12n+5}-\frac{1}{12n+7})$$$$=\frac{1}{12}\sum _{n⩾1}(\frac{1}{n-\frac{11}{12}}-\frac{1}{n-\frac{1}{12}})+\frac{1}{12}\sum _{n⩾1}(\frac{1}{n-\frac{7}{12}}-\frac{1}{n-\frac{5}{12}})$$$$=\frac{1}{12}(\psi \left(\frac{11}{12}\right)-\psi \left(\frac{1}{12}\right))+\frac{1}{12}(\psi \left(\frac{7}{12}\right)-\psi \left(\frac{5}{12}\right))$$$$weknowthat:$$$$\psi (1-s)-\psi \left(s\right)=\pi cot\left(\pi s\right)$$$$\therefore S=\frac{\pi }{12}(cot\left(\frac{\pi }{12}\right)+cot\left(\frac{5\pi }{12}\right))$$$$note::tan\left({15}^{°}\right)=\frac{1-\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}=\frac{12-6\sqrt{3}}{6}=2-\sqrt{3}$$$$S=\frac{\pi }{12}(2+\sqrt{3}+2-\sqrt{3})=\frac{\pi }{3}$$$$i{don}^{\prime }tknowfinallanswer$$$$iscorrectornopleasetellme.$$$$$$

Commented by Dwaipayan Shikari last updated on 12/Nov/20

$$Yesitiscorrectsir!$$

Commented by mnjuly1970 last updated on 12/Nov/20

$$thankyousomuch$$$$yourquestionsisverynice$$$$grateful....$$

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