Advanced-calculus-Find-the-value-of-the-following-series-n-1-1-n-n-n-2-n-z-n-1-1-n-z-

Question Number 188381 by mnjuly1970 last updated on 28/Feb/23

Advanced calculus

Find the value of the following series .

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

ζ (z) = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{z}}; R e (z) > 1

Answered by witcher3 last updated on 15/Mar/23

n ⩾ 2

ζ (1) diverge

Answered by witcher3 last updated on 18/Mar/23

Ω = \sum_{n ⩾ 2} \frac{{(- 1)}^{n} ζ (n)}{n . 2^{n}}

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

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

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

\sum_{n ⩾ 0} \frac{{(- 1)}^{n} x^{n - 1}}{2^{n} . n!} cv Uniformly to \exp

this We can exchange Σ and \int in Ω

\Rightarrow Ω = \int_{0}^{\infty} \frac{1}{x} \sum_{n ⩾ 2} \frac{{(- \frac{x}{2})}^{n}}{n!} dx = \int_{0}^{\infty} \frac{e^{- \frac{x}{2}} - 1 + \frac{x}{2}}{x (e^{x} - 1)} dx

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

f (a) = \int_{0}^{\infty} e^{- ax} (- \frac{1}{x (e^{- \frac{x}{2}} + 1)} + \frac{1}{2 (1 - e^{- x})}) dx

f^{'} (a) = \int_{0}^{\infty} e^{- ax} (\frac{1}{e^{- \frac{x}{2}} + 1} - \frac{x}{2 (1 - e^{- x})}) dx

\int_{0}^{\infty} - \frac{x}{2 (1 - e^{- x})} e^{- ax} dx = - \frac{1}{2} \sum_{k ⩾ 0} \int_{0}^{\infty} {xe}^{- (a + k) x}

= - \frac{1}{2} . Γ (2) \sum_{k ⩾ 0} \frac{1}{{(k + a)}^{2}} = - \frac{1}{2} Ψ^{(1)} (a)

\int_{0}^{\infty} \frac{e^{- ax}}{e^{- \frac{x}{2}} + 1} = \sum_{k ⩾ 0} {(- 1)}^{k} \int_{0}^{\infty} e^{- (a + \frac{k}{2}) x}

= \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(a + \frac{k}{2})} = \sum_{k ⩾ 0} \frac{1}{a + k} - \frac{1}{a + \frac{1}{2} + k}

= Ψ (a + \frac{1}{2}) - Ψ (a)

f^{'} (a) = Ψ (a + \frac{1}{2}) - Ψ (a) - \frac{Ψ^{(1)} (a)}{2}

f (a) = \log (\frac{Γ (a + \frac{1}{2})}{Γ (a)}) - \frac{1}{2} Ψ (a) + c

\lim_{a \to \infty} \int_{0}^{\infty} e^{- ax} (- \frac{1}{x (e^{- \frac{x}{2}} + 1)} + \frac{1}{2 (1 - e^{- x})}) dx

= \int_{0}^{\infty} {\underset{a \to \infty}{\lim e}}^{- ax} (- \frac{1}{x (e^{- \frac{x}{2}} + 1)} + \frac{1}{2 (1 - e^{- x})}) dx = 0

\Rightarrow \underset{a \to \infty}{\lim l n} (\frac{Γ (a + \frac{1}{2})}{Γ (a)}) - \frac{Ψ (a)}{2} + c = 0

Γ (a) \sim \sqrt{2 π} . a^{a - \frac{1}{2}} e^{- a}

Γ (a + \frac{1}{2}) \sim \sqrt{2 π} a^{a} e^{- a}

Ψ (a) \sim \ln (a)

\underset{a \to \infty}{\lim f} (a) = \underset{a \to \infty}{\lim l n} (\sqrt{a}) - \frac{1}{2} (\ln (a)) = 0 \Rightarrow c = 0

f (a) = \ln (\frac{Γ (a + \frac{1}{2})}{Γ (a)}) - \frac{Ψ (a)}{2}

Ω = f (1) = \ln (\frac{Γ (\frac{3}{2})}{Γ (1)}) - \frac{Ψ (1)}{2} = \ln (\frac{\sqrt{π}}{2}) - \frac{Ψ (1)}{2}

= \frac{\ln (π) + γ}{2} - \ln (2) = \sum_{n ⩾ 2} \frac{{(- 1)}^{n} ζ (n)}{2^{n} . n}

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