prove-n-1-9n-4-3n-3n-1-3n-2-3-2-ln-3-

Question Number 97956 by M±th+et+s last updated on 10/Jun/20

p r o v e

\underset{n = 1}{\sum^{\infty}} \frac{9 n + 4}{3 n (3 n + 1) (3 n + 2)} = \frac{3}{2} - l n (3)

Answered by maths mind last updated on 13/Jun/20

= \sum_{n ⩾ 1} \frac{4 (3 n + 1) - 3 n}{3 n (3 n + 1) (3 n + 2)}

= 4 Σ \frac{1}{3 n (3 n + 2)} - Σ \frac{1}{(3 n + 1) (3 n + 2)}

= \frac{4}{9} \sum_{n ⩾ 1} \frac{1}{n (n + \frac{2}{3})} - \frac{1}{9} \sum_{n ⩾ 1} \frac{1}{(n + \frac{1}{3}) (n + \frac{2}{3})}

= \frac{4}{9} \sum_{n ⩾ 0} \frac{1}{(n + 1) (n + \frac{5}{3})} - \frac{1}{9} \sum_{n ⩾ 0} \frac{1}{(n + \frac{4}{3}) (n + \frac{5}{3})}

= \frac{4}{9} . \frac{Ψ (\frac{5}{3}) - Ψ (1)}{\frac{5}{3} - 1} - \frac{1}{9} . \frac{Ψ (\frac{5}{3}) - Ψ (\frac{4}{3})}{\frac{5}{3} - \frac{4}{3}}

S = \frac{2}{3} (Ψ (\frac{5}{3}) - Ψ (1)) - \frac{1}{3} (Ψ (\frac{5}{3}) - Ψ (\frac{4}{3}))

\frac{1}{3} (Ψ (\frac{5}{3}) + Ψ (\frac{4}{3})) - \frac{2}{3} Ψ (1)

Ψ (\frac{p}{q}) = - γ - l n (2 q) - \frac{π}{2} c o t (\frac{p π}{q}) + 2 \underset{k = 0}{\sum^{[\frac{q - 1}{2}]}} c o s (\frac{2 p π k}{q}) l n (s i n (\frac{p k π}{q}))

Ψ (\frac{5}{3}) = \frac{3}{2} + Ψ (\frac{2}{3}), Ψ (\frac{4}{3}) = 3 + Ψ (\frac{1}{3})

Ψ (\frac{2}{3}) = - γ - l n (6) - \frac{π}{2} c o t (\frac{2 π}{3}) + 2 c o s (\frac{4 π}{3}) l n (s i n (\frac{2 π}{3}))

= - γ - l n (6) + \frac{π}{2 \sqrt{3}} - l n (\frac{\sqrt{3}}{2})

Ψ (\frac{1}{3}) = - γ - l n (6) - \frac{π}{2 \sqrt{3}} - l n (\frac{\sqrt{3}}{2})

S = \frac{1}{3} (- 2 γ - l n (27) + \frac{3}{2} + 3) - \frac{2}{3} (- γ)

S = l n (\frac{1}{3}) + \frac{3}{2} = \frac{3}{2} - l n (3)

Commented by M±th+et+s last updated on 13/Jun/20

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