challanging-math-prove-that-n-2-1-n-n-2-n-1-n-ln-pi-4-hint-s-1-s-0-x-s-1-e-x-1-dx-m-n-

Question Number 122185 by mnjuly1970 last updated on 14/Nov/20

\dots c h a l l a n g i n g m a t h \dots

p r o v e t h a t ::

\underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{2^{n - 1} n} \overset{? ? ?}{=} γ + l n (\frac{π}{4})

h i n t :: ζ (s) = \frac{1}{Γ (s)} \int_{0}^{\infty} \frac{x^{s - 1}}{e^{x} - 1} d x

m . n \dots

Commented by Dwaipayan Shikari last updated on 15/Nov/20

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

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

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

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

\underset{s = 1}{\sum^{\infty}} \frac{1}{s} + 2 \sum^{\infty} l o g (1 + \frac{1}{2 s})

l o g (Γ (\frac{1}{2})) = - \frac{γ}{2} + l o g 2 + \sum^{\infty} \frac{1}{2 s} + \sum^{\infty} l o g (1 + \frac{1}{2 s})

l o g (π) = - γ + l o g 4 + \sum^{\infty} \frac{1}{s} + 2 \sum^{\infty} l o g (1 + \frac{1}{2 s})

\sum^{\infty} \frac{1}{s} + 2 \sum^{\infty} l o g (1 + \frac{1}{2 s}) = γ + l o g (\frac{π}{4})

Commented by mnjuly1970 last updated on 15/Nov/20

v e r y n i c e . b r a v o

m a s t e r d w a i p a y a n . .

Answered by mindispower last updated on 15/Nov/20

\sum_{n ⩾ 2} \frac{{(- 1)}^{n} x^{n} ζ (n)}{n}

= \sum_{n ⩾ 2} \frac{{(- x)}^{n}}{n} . \sum_{k ⩾ 1} \frac{1}{k^{n}}

= \sum_{k ⩾ 1} \sum_{n ⩾ 2} \frac{{(- \frac{x}{k})}^{n}}{n} = S (x)

= \sum_{k ⩾ 1} - (l n (1 + \frac{x}{k}) - \frac{x}{k}) = S (x)

Γ (x) = \frac{e^{- γ x}}{x} \prod_{k ⩾ 1} \frac{e^{\frac{x}{k}}}{1 + \frac{x}{k}} \Rightarrow

l o g (Γ (z)) - γ x - l n (x) + \sum_{k ⩾ 1} (\frac{x}{k} - l n (1 + \frac{x}{k}))

\Rightarrow S (x) = l o g (Γ (x)) + γ x + l n (x)

= S (x) = l o g (x Γ (x)) + γ x = l o g (Γ (x + 1)) + γ x

Σ \frac{{(- 1)}^{n} ζ (n)}{2^{n - 1} n} = 2 \sum_{n ⩾ 2} \frac{{(- 1)}^{n} {(\frac{1}{2})}^{n} ζ (n)}{n} = 2 S (\frac{1}{2})

2 (l o g (Γ (\frac{1}{2}) + \frac{γ}{2}) = l o g \frac{π}{4} + γ

w i t h e h i n t l e a d s t o

\int_{0}^{\infty} \frac{2}{x} (e^{- \frac{x}{2}} - 1 + \frac{x}{2}) . \frac{d x}{e^{x} - 1}

Commented by mnjuly1970 last updated on 15/Nov/20

p e a c e b e u p o n y o u s i r

m i n d s p o w e r . r e a l l y n i c e a n d

e x c e l l e n t .

Answered by mnjuly1970 last updated on 15/Nov/20

Answered by mnjuly1970 last updated on 15/Nov/20