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Question Number 211558 by mnjuly1970 last updated on 12/Sep/24

$$$$$$$$$$------------$$$$\mathbf{\Omega }=\underset{\mathit{n}=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{3\mathit{n}+2}-\frac{1}{3\mathit{n}+1})=\mathit{a}\mathit{\pi }$$$$\Rightarrow {\mathit{a}}^2=?$$$$------------$$

Answered by mr W last updated on 29/Sep/24

$$\frac{1}{3}\psi (1+\frac{2}{3})=-\frac{\gamma }{3}+\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n}-\frac{1}{3n+2})$$$$\frac{1}{3}\psi (1+\frac{1}{3})=-\frac{\gamma }{3}+\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n}-\frac{1}{3n+1})$$$$\frac{1}{3}[\psi (1+\frac{1}{3})-\psi (1+\frac{2}{3})]=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n+2}-\frac{1}{3n+1})$$$$\frac{1}{3}[\psi (1+\frac{1}{3})-\psi (1+\frac{2}{3})]=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n+2}-\frac{1}{3n+1})+\frac{1}{2}$$$$\frac{1}{3}[\psi (1+\frac{1}{3})-\psi (1+\frac{2}{3})]-\frac{1}{2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n+2}-\frac{1}{3n+1})$$$$\frac{1}{3}[\psi \left(\frac{1}{3}\right)+3-\psi \left(\frac{2}{3}\right)-\frac{3}{2}]-\frac{1}{2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n+2}-\frac{1}{3n+1})$$$$\frac{1}{3}[\psi \left(\frac{1}{3}\right)-\psi (1-\frac{1}{3})]=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n+2}-\frac{1}{3n+1})$$$$\frac{1}{3}[\psi \left(\frac{1}{3}\right)-\psi \left(\frac{1}{3}\right)-\pi \cot \frac{\pi }{3}]=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n+2}-\frac{1}{3n+1})$$$$-\frac{\pi }{3\sqrt{3}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{3n+2}-\frac{1}{3n+1})=a\pi$$$$\Rightarrow a=-\frac{1}{3\sqrt{3}}$$$$\Rightarrow a^2=\frac{1}{27}✓$$

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