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Question Number 157120 by cortano last updated on 20/Oct/21

\underset{n = 0}{\sum^{\infty}} (\frac{1}{(3 n + 1) (3 n + 2)}) = ?

Answered by puissant last updated on 20/Oct/21

Ω = \underset{n = 0}{\sum^{\infty}} (\frac{1}{(3 n + 1) (3 n + 2)}) = \frac{1}{9} \underset{n = 0}{\sum^{\infty}} \frac{1}{(n + \frac{1}{3}) (n + \frac{2}{3})}

= \frac{1}{9} {\frac{ψ (\frac{2}{3}) - ψ (\frac{1}{3})}{\frac{2}{3} - \frac{1}{3}}} = \frac{1}{3} {ψ (\frac{2}{3}) - ψ (\frac{1}{3})}

∙ ψ (1 - z) - ψ (z) = π c o t a n (π z) f o r z = \frac{1}{3}

\to ψ (\frac{2}{3}) = ψ (\frac{1}{3}) + π c o t a n (\frac{π}{3}) = ψ (\frac{1}{3}) + \frac{π}{\sqrt{3}} . .

\Rightarrow Ω = \frac{1}{3} {ψ (\frac{1}{3}) + \frac{π}{\sqrt{3}} - ψ (\frac{1}{3})} = \frac{π}{3 \sqrt{3}} . .

∴∵ Ω = \underset{n = 0}{\sum^{\infty}} (\frac{1}{(3 n + 1) (3 n + 2)}) = \frac{π}{3 \sqrt{3}} . .

\dots \dots \dots \dots \dots L e p u i s s a n t \dots \dots \dots \dots \dots

Commented by Tawa11 last updated on 20/Oct/21

great sir

Answered by cortano last updated on 20/Oct/21

\underset{n = 0}{\sum^{\infty}} (\frac{1}{(3 n + 1) (3 n + 2)})

= \underset{n = 0}{\sum^{\infty}} \underset{0}{\int^{1}} (x^{3 n} - x^{3 n + 1}) d x

= \underset{0}{\int^{1}} (1 - x) \underset{n = 0}{\sum^{\infty}} x^{3 n} d x

= \underset{0}{\int^{1}} (\frac{1 - x}{1 - x^{3}}) d x = \underset{0}{\int^{1}} (\frac{d x}{1 + x + x^{2}})

= \frac{π}{3 \sqrt{3}}

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