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Question Number 132715 by Dwaipayan Shikari last updated on 16/Feb/21

\frac{1}{1^{3}} - \frac{1}{2^{3}} + \frac{1}{4^{3}} - \frac{1}{5^{3}} + \frac{1}{7^{3}} - \frac{1}{8^{3}} + \dots

Answered by Olaf last updated on 16/Feb/21

Let {(a_{n})}_{n \in N} = {1, 2, 4, 5, 7, 8 \dots}

a_{n} = \frac{1}{4} [6 n + 3 + {(- 1)}^{n}]

S = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{a_{n}^{3}}

S = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} 4^{3}}{{[6 n + 3 + {(- 1)}^{n}]}^{3}}

S = \underset{p = 0}{\sum^{\infty}} \frac{{(- 1)}^{2 p} 4^{3}}{{[6 (2 p) + 3 + {(- 1)}^{2 p}]}^{3}}

+ \underset{p = 0}{\sum^{\infty}} \frac{{(- 1)}^{2 p + 1} 4^{3}}{{[6 (2 p + 1) + 3 + {(- 1)}^{2 p + 1}]}^{3}}

S = \underset{p = 0}{\sum^{\infty}} \frac{4^{3}}{{(12 p + 4)}^{3}} - \underset{p = 0}{\sum^{\infty}} \frac{4^{3}}{{(12 p + 8)}^{3}}

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

S = - \frac{Ψ (2, \frac{1}{3})}{54} + \frac{Ψ (2, \frac{2}{3})}{54}

S = \frac{4 π^{3} \sqrt{3}}{243}

S \approx 0, 8840238122

Commented by Dwaipayan Shikari last updated on 16/Feb/21

G r e a t s i r!

Commented by Dwaipayan Shikari last updated on 16/Feb/21

\frac{π}{2} t a n (\frac{π}{2} x) = \frac{1}{1 - x} - \frac{1}{1 + x} + \frac{1}{3 - x} - \frac{1}{3 + x} + \dots

\frac{π^{2}}{4} {s e c}^{2} (\frac{π}{2} x) = \frac{1}{{(1 - x)}^{2}} + \frac{1}{{(1 + x)}^{2}} + \frac{1}{{(3 - x)}^{2}} + \frac{1}{{(3 + x)}^{2}} + \dots

\frac{π^{3}}{8} {s e c}^{2} (\frac{π}{2} x) t a n (\frac{π}{2} x) = \frac{1}{{(1 - x)}^{3}} - \frac{1}{{(1 + x)}^{3}} + \dots

x = \frac{1}{3}

\frac{π^{3}}{8} . \frac{4}{3} . \frac{1}{\sqrt{3}} = \frac{3^{3}}{2^{3}} - \frac{3^{3}}{4^{3}} + \frac{3^{3}}{8^{3}} - \frac{3^{3}}{10^{3}} + \dots

= \frac{π^{3}}{27.6 \sqrt{3}} = \frac{1}{2^{3}} (1 - \frac{1}{2^{3}} + \frac{1}{4^{3}} - \frac{1}{5^{3}} + \dots)

\Rightarrow \frac{4 π^{3}}{81 \sqrt{3}} = 1 - \frac{1}{2^{3}} + \frac{1}{4^{3}} - \frac{1}{5^{3}} + \dots

Commented by mnjuly1970 last updated on 16/Feb/21

p l e a s e p r o v e t h i s n i c e i d e n t i t y

m r d w a i p a y a n . .

\frac{π}{2} t a n (\frac{π}{2} (x)) = \frac{1}{1 - x} - \frac{1}{1 + x} + \frac{1}{3 + x} - \frac{1}{3 - x} \dots

o f c o u r s e i t h i n k

i t c a n b e d e d u c e d b y

w e i r e s t e r e s s g a m m a f o r m u l a . .

Γ (x) = \frac{e^{- γ x}}{x} \underset{k = 1}{\prod^{\infty}} \frac{e^{\frac{x}{k}}}{1 + \frac{x}{k}}

a n d i t s c o r a l l a r y

\frac{s i n (π x)}{π x} = \underset{k = 1}{\prod^{\infty}} (1 - \frac{x^{2}}{k^{2}}) . .

Commented by Dwaipayan Shikari last updated on 16/Feb/21

c o s x = (1 - \frac{2 x}{π}) (1 + \frac{2 x}{π}) (1 - \frac{2 x}{3 π}) (1 + \frac{2 x}{3 π}) (1 - \frac{2 x}{5 π}) (1 + \frac{2 x}{5 π}) \dots

l o g (c o s x) = l o g (1 - \frac{2 x}{π}) + l o g (1 + \frac{2 x}{π}) + \dots

D i f f e r e n t i a t i n g b o t h s i d e s r e s p e c t t o x

- t a n x = \frac{\frac{- 2}{π}}{1 - \frac{2 x}{π}} + \frac{\frac{2}{π}}{1 + \frac{2 x}{π}} - \frac{\frac{2}{3 π}}{1 - \frac{2 x}{3 π}} + . .

t a n x = \frac{1}{\frac{π}{2} - x} - \frac{1}{\frac{π}{2} + x} + \frac{1}{\frac{3 π}{2} - x} - \frac{1}{\frac{3 π}{2} + x} + \dots

\frac{π}{2} t a n (\frac{π}{2} x) = \frac{1}{1 - x} - \frac{1}{1 + x} + \frac{1}{3 - x} - \frac{1}{3 + x} + \frac{1}{5 - x} - \frac{1}{5 + x} + \dots

Commented by mnjuly1970 last updated on 16/Feb/21

g r a t e f u l m r D w a i p a y a n

t h a n k s l o t \dots

Answered by mnjuly1970 last updated on 16/Feb/21

s o l u t i o n :

S i s c o n v e r g e n t \dots

r e a r r a n g e m e n t \dots

S = (\frac{1}{1^{3}} + \frac{1}{4^{3}} + \frac{1}{7^{3}} + \dots .) +

(\frac{1}{2^{3}} + \frac{1}{5^{3}} + \frac{1}{8^{3}} + \dots)

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

= \frac{1}{27} {\underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{1}{3})}^{3}} - \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{2}{3})}^{3}}}

n o t e :: (1) :: ψ (z) - ψ (1 - z) = - π c o t (π z)

(2) : : ψ^{'} (z) + ψ^{'} (1 - z) = π^{2} (1 + {c o t}^{2} (π z))

(3) :: ψ^{″} (z) - ψ^{″} (1 - z) = - 2 π^{3} (1 + {c o t}^{2} (π z)) c o t (π z)

ψ^{″} (\frac{1}{3}) - ψ^{″} (\frac{2}{3}) = - 2 π^{3} (1 + \frac{1}{3}) (\frac{1}{\sqrt{3}})

ψ^{″} (\frac{1}{3}) - ψ^{″} (\frac{2}{3}) = \frac{- 8 π^{3}}{3 \sqrt{3}} \dots .

S = \frac{1}{27} (\frac{- ψ^{″} (\frac{1}{3})}{2} + \frac{ψ^{″} (\frac{2}{3})}{2})

= \frac{1}{54} (\frac{8 π^{3}}{3 \sqrt{3}}) = \frac{4 π^{3}}{81 \sqrt{3}} \dots . ✓ ✓ m . n . .

Answered by Ajetunmobi last updated on 16/Feb/21

let the sum = S

seperating the positive term and negative term

and factor out minus we get

S = \underset{k = 0}{\sum^{\infty}} \frac{1}{{(3 k + 1)}^{3}} - \sum_{k = 0} \frac{1}{{(3 k + 2)}^{3}} = \frac{1}{27} {\sum_{k = 0} \frac{1}{{(k + \frac{1}{3})}^{3}} - \sum_{k = 0} \frac{1}{{(k + \frac{2}{3})}^{3}}}

S = \frac{1}{27} {ψ^{(1)} (\frac{1}{3}) - ψ^{(1)} (\frac{2}{3})} Answer

Ajetunmobi Abdulqoyyum

Answered by Ajetunmobi last updated on 16/Feb/21

it should be tetragamma not trigamma

sorry for the mistakes