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Question Number 122467 by Dwaipayan Shikari last updated on 17/Nov/20

\frac{1}{π^{2} - 1} + \frac{1}{4 π^{2} - 1} + \frac{1}{9 π^{2} - 1} + \frac{1}{16 π^{2} - 1} + \dots .

Commented by Dwaipayan Shikari last updated on 17/Nov/20

c o t x = \frac{1}{x} - \frac{1}{π - x} + \frac{1}{π + x} - \frac{1}{2 π - x} + \frac{1}{2 π + x} - \dots

c o t (1) = 1 - \frac{2}{π^{2} - 1} - \frac{2}{4 π^{2} - 1} - \frac{2}{9 π^{2} - 1} - \dots

1 - c o t (1) = 2 (\frac{1}{π^{2} - 1} + \frac{1}{4 π^{2} - 1} + \frac{1}{9 π^{2} - 1} + \dots)

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} π^{2} - 1} = \frac{1 - c o t (1)}{2}

Answered by mnjuly1970 last updated on 17/Nov/20

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

l n (s i n (π x)) - l n (π x) = \underset{k = 1}{\sum^{\infty}} l o g (1 - \frac{x^{2}}{k^{2}})

∴ π c o t (π x) - \frac{1}{x} = \underset{k = 1}{\sum^{\infty}} (\frac{\frac{- 2 x}{k^{2}}}{1 - \frac{x^{2}}{k^{2}}})

∴ π c o t (π x) - \frac{1}{x} = - 2 x \underset{k = 1}{\sum^{\infty}} (\frac{1}{k^{2} - x^{2}})

∴ \frac{1}{2 x^{2}} - \frac{π}{2 x} c o t (π x) = \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{2} - x^{2}} ✓

x = \frac{1}{π} \Rightarrow \frac{π^{2}}{2} - \frac{π^{2}}{2} c o t (1) = \underset{k = 1}{\sum^{\infty}} \frac{π^{2}}{k^{2} π^{2} - 1}

∴ Σ \frac{1}{k^{2} π^{2} - 1} = \frac{1}{2} (1 - c o t (1)) ✓

\dots c o r r e c t o r n o ?

m r . D w a i p a y a n \dots

∴ . m . n . 1970 .

Commented by Dwaipayan Shikari last updated on 17/Nov/20

E r r o r i n 4 t h l i n e x = \frac{1}{π}

S o i t b e c o m e s \frac{π^{2}}{2} - \frac{π^{2} c o t (1)}{2} = π^{2} \sum^{\infty} \frac{1}{k^{2} π^{2} - 1}

\sum^{\infty} \frac{1}{k^{2} π^{2} - 1} = \frac{1}{2} (1 - c o t (1))

Commented by Dwaipayan Shikari last updated on 17/Nov/20

T h a n k i n g y o u f o r I n t e r a c t i n g w i t h t h i s P r o b l e m

Commented by mnjuly1970 last updated on 17/Nov/20

t h a n k y o u s o m u c h \dots

Commented by mnjuly1970 last updated on 17/Nov/20

e x c u s e m e m r p a y a n . b u t

i c o u l d e n t f i n d a n y e r r o r

i n 4 t h l i n e . i d i f f e r e n t i a t e d

b o t h s i d e s a n d s i m p l i f i e d i t . \dots

Commented by mnjuly1970 last updated on 17/Nov/20

i a m c h e c k i n g i t .

Commented by mnjuly1970 last updated on 17/Nov/20

p l e a s e r e v i e w a n d r e c h e c k .

Commented by Dwaipayan Shikari last updated on 17/Nov/20

Y o u h a v e e d i t e d m a n y t i m e s a g o . t h e r e i s n o m i s t a k e n o w

s i r ? :)

Y o u a r e w o n d e r i n g a b o u t t h e m i s t a k e

B u t a c t u a l l y t h e r e i s n o m i s t a k e :)

Commented by mnjuly1970 last updated on 17/Nov/20

y o u a r e a l s o f u n n y p e r s o n . j o k e i s a l s o a n a s p e c t

o f m a t h e m a t i c s .

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

t h a n m e .

b y t h e w a y, y o u h a v e a g o o d f u t u r e

i n m a t h . g o o d l u c k y o u n g m a n .

Commented by Dwaipayan Shikari last updated on 17/Nov/20

M y a g e i s e^{2.82}

Commented by mnjuly1970 last updated on 18/Nov/20

m e r c e y . \approx 17

m y a g e i s \approx 3 e^{2.82}