Advanced-Calculus-if-n-2-1-n-n-2-n-then-prove-that-1-2-e-1-proof-method-1-

Question Number 141668 by mnjuly1970 last updated on 22/May/21

\dots \dots . A d v a n c e d \dots ★ \dots ★ \dots C a l c u l u s \dots \dots .

i f Ω = \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{2^{n}} t h e n p r o v e

t h a t :: \frac{1}{2} = e^{Ω - 1}

p r o o f ::

m e t h o d (1) :

ψ (1 + x) = - γ + \underset{n = 2}{\sum^{\infty}} {(- 1)}^{n} ζ (n) x^{n - 1}

(M a c l a u r i n s e r i e s f o r ψ (x + 1))

x := \frac{1}{2} \Rightarrow ψ (\frac{3}{2}) = - γ + 2 \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{2^{n}} (*)

w e k n o w t h a t :: ψ (1 + x) = \frac{1}{x} + ψ (x)

(*) ⇛ ψ (\frac{3}{2}) = 2 + ψ (\frac{1}{2}) = - γ + 2 \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{2^{n}}

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

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

\frac{1}{2} = e^{Ω - 1} \dots . ✓

\dots m . n . j u l y . 1970 \dots

Answered by Dwaipayan Shikari last updated on 22/May/21

ψ (z + 1) = - γ + \underset{n = 2}{\sum^{\infty}} {(- 1)}^{n} ζ (n) z^{n - 1}

z^{2} ψ (z + 1) = - γ z^{2} + \underset{n = 2}{\sum^{\infty}} {(- 1)}^{n} ζ (n) z^{n + 1} ψ (\frac{1}{2}) = - γ - l o g (4)

z = \frac{1}{2} \Rightarrow \frac{ψ (\frac{3}{2}) + γ}{4} = \underset{n = 2}{\sum^{\infty}} {(- 1)}^{n} ζ (n) \frac{1}{2^{n + 1}}

\Rightarrow \frac{l o g (e / 2)}{2} = \underset{n = 2}{\sum^{\infty}} {(- 1)}^{n} \frac{ζ (n)}{2^{n + 1}}

e^{Ω - 1} = \frac{1}{\sqrt{2}}

Commented by mnjuly1970 last updated on 22/May/21

t h a n k s a l o t m r p a y a n

i e d i t e d i t . . g r a t e f u l . .

p l e a s e r e c h e c k m y a n s w e r . .

Commented by Dwaipayan Shikari last updated on 22/May/21

Y e s s i r i t i s C o r r e c t