calculate-n-1-2n-2-n-n-

Question Number 201106 by mnjuly1970 last updated on 29/Nov/23

c a l c u l a t e \dots

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

Answered by witcher3 last updated on 30/Nov/23

ζ (2 n) Γ (2 n) = \int_{0}^{\infty} \frac{x^{2 n - 1}}{e^{x} - 1} dx

S = \sum_{n ⩾ 1} \frac{ζ (2 n)}{2^{n} n} = \sum_{n ⩾ 1} \frac{1}{2^{n} Γ (2 n) . n} \int_{0}^{\infty} \frac{x^{2 n - 1}}{e^{x} - 1} dx

= \int_{0}^{\infty} \frac{1}{x} \sum_{n ⩾ 1} [\frac{x^{2 n}}{(2 n)! . 2^{n}}] . \frac{dx}{e^{x} - 1}

= \int_{0}^{\infty} \frac{ch (\frac{x}{\sqrt{2}}) - 1}{x (e^{x} - 1)} dx = S

S (a) = \int_{0}^{\infty} \frac{ch (ax) - 1}{x (e^{x} - 1)} dx; S (0) = 0

s^{'} (a) = \int_{0}^{\infty} \frac{sh (ax)}{e^{x} - 1} dx

= \int_{0}^{\infty} \frac{e^{ax} - e^{- ax}}{e^{x} - 1} dx, e^{- x} = t

= \int_{0}^{1} \frac{t^{- a} - t^{a}}{1 - t} dt = Ψ (a + 1) - Ψ (1 - a)

= \frac{1}{a} - π \cot (π a)

s (a) = \ln (a) - \ln (\sin (π a) + c

= \ln (\frac{a}{\sin (π a)}) + c

\underset{a \to 0}{\lim s} (a) = s (0) = 0 = \ln (\frac{1}{π}) + c \Rightarrow c = \ln (π)

S (a) = \ln (\frac{π a}{\sin (π a)})

S = s (\frac{1}{\sqrt{2}}) = \ln (\frac{π}{\sqrt{2} . \sin (\frac{π}{\sqrt{2}})})

Commented by mnjuly1970 last updated on 01/Dec/23

t h a n k s a l o t m a s t e r ⋚

Commented by witcher3 last updated on 01/Dec/23

withe pleasur Barak alah fik