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Question Number 214485 by depressiveshrek last updated on 09/Dec/24

$$\underset{n\to \mathrm{\infty }}{\lim }(\frac{1}{2\cdot 4}+\frac{1}{5\cdot 7}+...+\frac{1}{(3n-1)(3n+1)})$$

Answered by mr W last updated on 10/Dec/24

$$\psi (1-\frac{1}{3})=-\gamma +\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-\frac{1}{n-\frac{1}{3}})$$$$\psi (1+\frac{1}{3})=-\gamma +\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-\frac{1}{n+\frac{1}{3}})$$$$\psi (1+\frac{1}{3})-\psi (1-\frac{1}{3})=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n-\frac{1}{3}}-\frac{1}{n+\frac{1}{3}})$$$$\psi \left(\frac{1}{3}\right)+3-\psi \left(\frac{1}{3}\right)-\pi \cot \frac{\pi }{3}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n-\frac{1}{3}}-\frac{1}{n+\frac{1}{3}})$$$$3-\frac{\pi }{\sqrt{3}}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n-\frac{1}{3}}-\frac{1}{n+\frac{1}{3}})$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(3n-1)(3n+1)}$$$$=\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n-1}-\frac{1}{3n+1})$$$$=\frac{1}{6}\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n-\frac{1}{3}}-\frac{1}{n+\frac{1}{3}})$$$$=\frac{1}{6}(3-\frac{\pi }{\sqrt{3}})$$$$=\frac{1}{2}-\frac{\pi }{6\sqrt{3}}✓$$

Commented by MathematicalUser2357 last updated on 12/Dec/24

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