super-nice-show-that-6-pi-6-945-

Question Number 127020 by bramlexs22 last updated on 26/Dec/20

s u p e r n i c e!

s h o w t h a t

ζ (6) = \frac{π^{6}}{945}

Commented by liberty last updated on 26/Dec/20

h a h a h a v e r y n i c e

Answered by Olaf last updated on 26/Dec/20

ζ (2 k) = \frac{{(- 1)}^{k - 1} B_{2 k} {(2 π)}^{2 k}}{2 (2 k)!}

For k = 3 :

ζ (6) = \frac{{(- 1)}^{2} B_{6} {(2 π)}^{6}}{2 \times 6!} = \frac{2}{45} B_{6} π^{6}

B_{6} = \frac{1}{42}

ζ (6) = \frac{2}{45} \times \frac{1}{42} π^{6} = \frac{π^{6}}{945}

Answered by Dwaipayan Shikari last updated on 26/Dec/20

ζ (2 n) = \frac{{(- 1)}^{n + 1} {(2 π)}^{2 n}}{2 (2 n)!} B_{2 n}

ζ (6) = \frac{{(2 π)}^{6}}{2.6!} . \frac{1}{42} = \frac{π^{6}}{15.63} = \frac{π^{6}}{945}

ζ (8) = \frac{π^{8}}{9450}

ζ (10) = \frac{π^{10}}{93555}

\dots .

Commented by Dwaipayan Shikari last updated on 26/Dec/20

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

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

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

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

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

\Rightarrow π x i + \frac{2 π x i}{e^{2 π x i} - 1} = 1 - 2 \underset{k ⩾ 1}{\sum^{\infty}} ζ (2 k) x^{2 k} \Rightarrow 1 - \underset{k ⩾ 0}{\sum^{\infty}} \frac{β_{2 k}}{2 (2 k!)} {(2 π i x)}^{2 k}

ζ (2 k) x^{2 k} = \frac{β_{2 k}}{2 (2 k!)} {(2 π i x)}^{2 k} \Rightarrow ζ (2 k) = {(- 1)}^{k + 1} \frac{β_{2 k} {(2 π)}^{2 k}}{2 (2 k!)}

Commented by mnjuly1970 last updated on 26/Dec/20

v e r y n i c e m r p a y a n

s o l u t i o n w i t h e x p l a n a t i o n . .

Answered by liberty last updated on 26/Dec/20

v i a F o u r i e r s i n e s e r i e s

b_{n} = \frac{2}{π} \int_{0}^{π} (π x - x^{2}) \sin (n x) d x = \frac{4}{π} . \frac{1 - {(- 1)}^{n + 1}}{n^{3}}

b_{n} = {\begin{cases} 0; i f n i s o d d \\ \frac{8}{π n^{3}}; i f n e v e n \end{cases}

w r i t t i n g n = 2 k - 1 f o r s o m e p o s i t i v e i n t e g e r k

w e g e t π x - x^{2} \sim \underset{k = 1}{\sum^{\infty}} \frac{8 \sin ((2 k - 1) x)}{π {(2 k - 1)}^{3}}

\frac{2}{π} \int_{0}^{π} {(π x - x^{2})}^{2} d x = \underset{k = 1}{\sum^{\infty}} {(\frac{8}{π {(2 k - 1)}^{3}})}^{2}

s i m p l i f y i n g t h i s y i e l d s

\frac{π^{4}}{15} = \frac{64}{π^{2}} \underset{k = 1}{\sum^{\infty}} {(\frac{1}{2 k - 1})}^{6} a n d \underset{k = 1}{\sum^{\infty}} \frac{1}{{(2 k - 1)}^{6}} = \frac{π^{6}}{960}

h o w e v e r

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

= \frac{63}{64} ζ (6)

ζ (6) = \frac{64}{63} . \frac{π^{6}}{960} = \frac{π^{6}}{945} .

Commented by mnjuly1970 last updated on 26/Dec/20

v e r y n i c e b r a v o m r l i b e r t y

e x c e l l e n t \dots