n-1-1-9n-2-3n-

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

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

Commented by EDWIN88 last updated on 15/May/21

= 1 - \ln \sqrt{3} - \frac{π}{6 \sqrt{3}}

S = \frac{1}{3} \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{3}{3 n + 1})

S = \frac{1}{3} \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + \frac{1}{3}})

S = \frac{1}{3} [ψ (\frac{4}{3}) + γ]

S = \frac{1}{3} [ψ (\frac{1}{3}) + 3 + γ]

S = \frac{1}{3} [3 - \frac{3}{2} \ln 3 - \frac{π}{2 \sqrt{3}}]

S = 1 - \frac{1}{2} \ln 3 - \frac{π}{6 \sqrt{3}}

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

\underset{n = 1}{\sum^{\infty}} \frac{1}{3 n (3 n + 1)} = \underset{n = 1}{\sum^{\infty}} \frac{1}{3 n} - \frac{1}{3 n + 1}

= \frac{1}{3} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \frac{1}{n + \frac{1}{3}} = \frac{1}{3} ψ (\frac{4}{3}) + \frac{γ}{3}