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Question Number 141030 by iloveisrael last updated on 15/May/21

$$\to ⟨\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{9n^2+3n}=?⟩\leftarrow$$

Commented by EDWIN88 last updated on 15/May/21

$$=1-\ln \sqrt{3}-\frac{\pi }{6\sqrt{3}}$$$$\mathrm{S}=\frac{1}{3}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{n}}-\frac{3}{3\mathrm{n}+1})$$$$\mathrm{S}=\frac{1}{3}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+\frac{1}{3}})$$$$\mathrm{S}=\frac{1}{3}[\psi \left(\frac{4}{3}\right)+\gamma ]$$$$\mathrm{S}=\frac{1}{3}[\psi \left(\frac{1}{3}\right)+3+\gamma ]$$$$\mathrm{S}=\frac{1}{3}[3-\frac{3}{2}\ln 3-\frac{\pi }{2\sqrt{3}}]$$$$\mathrm{S}=1-\frac{1}{2}\ln 3-\frac{\pi }{6\sqrt{3}}$$$$$$

Answered by Dwaipayan Shikari last updated on 15/May/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{3n(3n+1)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{3n}-\frac{1}{3n+1}$$$$=\frac{1}{3}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{n+\frac{1}{3}}=\frac{1}{3}\psi \left(\frac{4}{3}\right)+\frac{\gamma }{3}$$

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