Evaluate-F-n-2-1-n-n-n-1-

Question Number 141328 by mnjuly1970 last updated on 17/May/21

\dots \dots E v a l u a t e :

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

\dots \dots .

Answered by Dwaipayan Shikari last updated on 17/May/21

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

\int_{0}^{1} z ψ (z + 1) = - \int_{0}^{1} γ z d z + \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{n + 1}

\Rightarrow {[l o g Γ (z + 1)]}_{0}^{1} - \int_{0}^{1} l o g (Γ (z + 1)) + \frac{γ}{2} = \underset{n = 2}{\sum^{\infty}} {(- 1)}^{n} \frac{ζ (n)}{n + 1}

\Rightarrow \int_{0}^{1} l o g (Γ (z)) + l o g (z) d z + \frac{γ}{2} = \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1} ζ (n)

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

Commented by mnjuly1970 last updated on 17/May/21

g r a t e f u l m r p a y a n .

v e r y n i c e \dots . t h a n k y o u f o r y o u r

c o n s t a n c o o p e r a t i o n \dots

Commented by Dwaipayan Shikari last updated on 17/May/21

T h a n k s s i r

Answered by mnjuly1970 last updated on 17/May/21

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

d i f f b o t h s i d e s r e s p e c t t o^{″} x^{″}

ψ (x + 1) = - γ + Σ {(- 1)}^{n} x^{n - 1} ζ (n)

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

\int_{0}^{1} x (\frac{1}{x} + ψ (x)) d x = - \frac{γ^{2}}{2} + Σ {(- 1)}^{n} ζ (n) . \frac{1}{n + 1}

1 + {[x l n (Γ (x))]}_{0}^{1} - \int_{0}^{1} l n (Γ (x)) d x = \frac{- γ^{2}}{2} + F

1 + {l i m}_{x \to 0^{+}} [(x l n (Γ (x))) \overset{?}{=} 0] - \frac{1}{2} l n (2 π)) + \frac{γ^{2}}{2} = F

F := 1 + \frac{γ^{2}}{2} - l n \sqrt{2 π} \dots \dots . ✓ ✓

? :: {l i m}_{x \to 0^{+}} (x l n (\frac{Γ (x + 1)}{x})) =

= {l i m}_{x \to 0^{+}} x l n (Γ (x + 1)) - {l i m}_{x \to 0^{+}} x l n (x)

= 0