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Question Number 122518 by Dwaipayan Shikari last updated on 17/Nov/20

$$1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+...$$

Commented by Dwaipayan Shikari last updated on 17/Nov/20

$$\frac{1}{2}tanx=\frac{1}{\pi -2x}-\frac{1}{\pi +2x}+\frac{1}{3\pi -2x}-\frac{1}{3\pi +2x}+\frac{1}{5\pi -2x}-\frac{1}{5x+2\pi }+...$$$$\frac{\sqrt{3}}{2}=\frac{1}{\pi -\frac{2\pi }{3}}-\frac{1}{\pi +\frac{2\pi }{3}}+\frac{1}{3\pi -\frac{2\pi }{3}}-\frac{1}{3\pi +\frac{2\pi }{3}}+\frac{1}{5\pi -\frac{2\pi }{3}}-..$$$$\frac{1}{2\sqrt{3}}=\frac{1}{\pi }-\frac{1}{5\pi }+\frac{1}{7\pi }-\frac{1}{11\pi }+\frac{1}{13\pi }-...$$$$1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-..=\frac{\pi }{2\sqrt{3}}$$

Answered by mnjuly1970 last updated on 18/Nov/20

$$S=1+\frac{1}{6}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+\frac{1}{6})}-\frac{1}{6}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(n-\frac{1}{6})}$$$$=1+\frac{1}{6}[\psi \left(\frac{5}{6}\right)-\psi \left(\frac{7}{6}\right)]$$$$=1+\frac{1}{6}[\psi \left(\frac{5}{6}\right)-\psi \left(\frac{1}{6}\right)-6]$$$$=\frac{1}{6}[\psi (1-\frac{1}{6})-\psi \left(\frac{1}{6}\right)]$$$$=\frac{1}{6}\left(\pi cot\left(\frac{\pi }{6}\right)\right)=\frac{\pi }{6}\ast \sqrt{3}=\frac{\pi }{2\sqrt{3}}✓$$$$$$

Commented by Dwaipayan Shikari last updated on 18/Nov/20

$$Greatsir!$$$$Ihaveusedthis$$$$tanx=\frac{2}{\pi -2x}-\frac{2}{\pi +2x}+\frac{2}{3\pi -2x}-\frac{2}{3\pi +2x}+..$$

Commented by mnjuly1970 last updated on 18/Nov/20

$$extraordinarysirdwaipayan$$

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