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

.......Advanced ...★ ...★ ... Calculus....... if Ω =Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) then prove that :: (1/2) = e^(Ω−1) proof :: method (1): ψ (1+x )= −γ+Σ_(n=2) ^∞ (−1)^n ζ(n)x^(n−1) ( Maclaurin series for ψ(x+1) ) x:=(1/2) ⇒ ψ ((3/2) )=−γ + 2Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) (∗ ) we know that :: ψ(1+x)=(1/x)+ψ(x) ( ∗ ) ⇛ ψ ((3/2))=2+ψ((1/2))=−γ+2Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) (∗) ⇛ 2−γ−ln(4)=−γ+2Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) ln((e/2))= Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) =Ω (1/2) = e^(Ω −1) ....✓ ...m.n.july.1970...

$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:.......{Advanced}\:...\bigstar\:...\bigstar\:...\:{Calculus}....... \\ $$$$\:\:\:\:\:\:\:\:\:{if}\:\:\:\:\Omega\:=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} }\:{then}\:{prove} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{that}\:::\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:{e}^{\Omega−\mathrm{1}} \:\: \\ $$$$\:\:\:\:{proof}\::: \\ $$$$\:\:\:\:{method}\:\left(\mathrm{1}\right): \\ $$$$\:\:\:\:\:\psi\:\left(\mathrm{1}+{x}\:\right)=\:−\gamma+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right){x}^{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\left(\:{Maclaurin}\:{series}\:{for}\:\psi\left({x}+\mathrm{1}\right)\:\right) \\ $$$$\:\:\:\:{x}:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\psi\:\left(\frac{\mathrm{3}}{\mathrm{2}}\:\right)=−\gamma\:+\:\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} }\:\:\left(\ast\:\right) \\ $$$$\:\:\:\:{we}\:{know}\:{that}\:::\:\psi\left(\mathrm{1}+{x}\right)=\frac{\mathrm{1}}{{x}}+\psi\left({x}\right) \\ $$$$\:\:\:\:\:\:\left(\:\ast\:\right)\:\:\Rrightarrow\:\psi\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{2}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\gamma+\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} } \\ $$$$\:\:\:\:\:\:\:\left(\ast\right)\:\:\:\:\:\Rrightarrow\:\:\:\:\:\:\:\:\:\mathrm{2}−\gamma−{ln}\left(\mathrm{4}\right)=−\gamma+\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{{e}}{\mathrm{2}}\right)=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} }\:=\Omega \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:{e}^{\Omega\:−\mathrm{1}} \:\:\:....\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.{july}.\mathrm{1970}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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

ψ(z+1)=−γ+Σ_(n=2) ^∞ (−1)^n ζ(n)z^(n−1) z^2 ψ(z+1)=−γz^2 +Σ_(n=2) ^∞ (−1)^n ζ(n)z^(n+1) ψ((1/2))=−γ−log(4) z=(1/2) ⇒ ((ψ((3/2))+γ)/4)=Σ_(n=2) ^∞ (−1)^n ζ(n)(1/(2^(n+1) )) ⇒((log(e/2))/2)=Σ_(n=2) ^∞ (−1)^n ((ζ(n))/2^(n+1) ) e^(Ω−1) =(1/( (√2)))

$$\psi\left({z}+\mathrm{1}\right)=−\gamma+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right){z}^{{n}−\mathrm{1}} \\ $$$${z}^{\mathrm{2}} \psi\left({z}+\mathrm{1}\right)=−\gamma{z}^{\mathrm{2}} +\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right){z}^{{n}+\mathrm{1}} \:\:\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\gamma−{log}\left(\mathrm{4}\right) \\ $$$${z}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\Rightarrow\:\:\frac{\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\gamma}{\mathrm{4}}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} \:} \\ $$$$\Rightarrow\frac{{log}\left({e}/\mathrm{2}\right)}{\mathrm{2}}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\zeta\left({n}\right)}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$${e}^{\Omega−\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$

Commented by mnjuly1970 last updated on 22/May/21

thanks alot mr payan i edited it..grateful.. please recheck my answer..

$${thanks}\:{alot}\:{mr}\:{payan} \\ $$$${i}\:{edited}\:{it}..{grateful}.. \\ $$$${please}\:{recheck}\:{my}\:{answer}.. \\ $$

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

Yes sir it is Correct

$${Yes}\:{sir}\:{it}\:{is}\:{Correct} \\ $$

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