show-that-n-1-1-2n-1-3n-1-1-6-3-pi-9log-3-log-4096-

Question Number 83144 by M±th+et£s last updated on 28/Feb/20

s h o w t h a t

\underset{n = 1}{\sum^{\infty}} \frac{1}{(2 n - 1) (3 n - 1)} = \frac{1}{6} (\sqrt{3} π - 9 l o g (3) + l o g (4096))

Answered by mind is power last updated on 28/Feb/20

= \frac{1}{6} \underset{n = 1}{\sum^{+ \infty}} \frac{1}{(n - \frac{1}{2}) (n - \frac{1}{3})}

= \frac{1}{6} \underset{n = 0}{\sum^{+ \infty}} \frac{1}{(n + \frac{1}{2}) (n + \frac{2}{3})} = \frac{1}{6} . \frac{Ψ (\frac{2}{3}) - Ψ (\frac{1}{2})}{\frac{2}{3} - \frac{1}{2}}

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{π}{3})

Ψ (\frac{1}{2}) = - γ - l n (4)

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

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

= π \sqrt{3} + l n (2^{12}) - 3 l n (3) = π \sqrt{3} + l n (4096) - 9 l n (3)

\underset{n = 1}{\sum^{+ \infty}} \frac{1}{(2 n - 1) (3 n - 1)} = \frac{1}{6} . (π \sqrt{3} - 9 l n (3) + l n (4096))

Commented by mind is power last updated on 28/Feb/20

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

Commented by M±th+et£s last updated on 28/Feb/20

g o d b l e s s y o u s i r

Commented by msup trace by abdo last updated on 28/Feb/20

d e f i n e Φ s i r m i n d \dots .

Commented by mathmax by abdo last updated on 28/Feb/20

t h a n k y o u s i r m i n d

Commented by mind is power last updated on 28/Feb/20

w i t h e p l e a s u r s i r