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Question Number 202589 by mnjuly1970 last updated on 30/Dec/23
Ω = Σ_(k=1) ^∞ ((1/k) + 2ln(1−(1/(2k))))=?
$$ \\ $$$$\:\:\Omega\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}}\:+\:\mathrm{2}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)\right)=?\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 30/Dec/23
Yes sir you are right Y
$$\:{Yes}\:{sir}\:\:{you}\:{are}\:{right}\:\cancel{\underline{\underbrace{\boldsymbol{\mathfrak{Y}}}}} \\ $$
Commented by MrGHK last updated on 30/Dec/23
sir,dont you think it diverves ?
$$\boldsymbol{\mathrm{sir}},\boldsymbol{\mathrm{dont}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{think}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{diverves}}\:? \\ $$
Answered by witcher3 last updated on 30/Dec/23
(1/k)+2ln(1−(1/(2k)))∼(1/k)+2(−(1/(2k))+(1/(4k^2 ))+o((1/k^2 ))) ∼(1/(2k^2 ))+o((1/k^2 )) U_n =(1/(2n^2 )) ΣU_n cv⇒Ω existe Ω=lim_(N→∞) Σ_(k=1) ^N (1/k)+2ln(((2k−1)/(2k)))) Ω_N =H_N +2ln(Π_(k=1) ^N (((2k−1)/(2k))))=H_N +2ln(((Π_(k=1) ^N (k−(1/2)))/(Π_(k=1) ^N (k)))) =H_N +2ln(((Γ(N+(1/2)))/(Γ((1/2))Γ(N+1)))) =H_N +ln(((Γ(N+(1/2)))/(Γ(N).N)))−2ln(Γ((1/2))) =H_N −ln(N)+2ln(((Γ(N+(1/2)))/(Γ(N).(√N))))−2ln((√π)) Ω=lim_(N→∞) Ω_N =lim_(N→∞) (H_N −ln(N)−ln(π)) +lim_(N→∞) log(((Γ(N+(1/2)))/(Γ(N)(√N))))− lim_(x→0) H_x −ln(x)=γ Γ(N+(1/2))∼(√(2π))(N+(1/2))^(N+(1/2)) e^(−(N+(1/2))) Γ(N) ∼(√(2π))N^N e^(−N) Ω=γ−log(π)+lim_(N→∞) .2log((((√(2π))(N+(1/2))^(N+(1/2)) e^(−N−(1/2)) )/( (√(2π)).N^N e^(−N) .(√N)))) γ−log(π)+lim_(N→∞) .2log((1+(1/(2N)))^N .((√(N+(1/2)))/N).e^(−(1/2)) ) =γ−log(π) Σ_(k=1) ^∞ (1/k)+2log(1−(1/(2k)))=γ−log(π)
$$\:\frac{\mathrm{1}}{\mathrm{k}}+\mathrm{2ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}\right)\sim\frac{\mathrm{1}}{\mathrm{k}}+\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2k}}+\frac{\mathrm{1}}{\mathrm{4k}^{\mathrm{2}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right)\right) \\ $$$$\sim\frac{\mathrm{1}}{\mathrm{2k}^{\mathrm{2}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right) \\ $$$$\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }\:\:\:\Sigma\mathrm{U}_{\mathrm{n}} \:\mathrm{cv}\Rightarrow\Omega\:\mathrm{existe} \\ $$$$\left.\Omega=\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}+\mathrm{2ln}\left(\frac{\mathrm{2k}−\mathrm{1}}{\mathrm{2k}}\right)\right) \\ $$$$\Omega_{\mathrm{N}} =\mathrm{H}_{\mathrm{N}} +\mathrm{2ln}\left(\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{N}} {\prod}}\left(\frac{\mathrm{2k}−\mathrm{1}}{\mathrm{2k}}\right)\right)=\mathrm{H}_{\mathrm{N}} +\mathrm{2ln}\left(\frac{\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{N}} {\prod}}\left(\mathrm{k}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{N}} {\prod}}\left(\mathrm{k}\right)}\right) \\ $$$$=\mathrm{H}_{\mathrm{N}} +\mathrm{2ln}\left(\frac{\Gamma\left(\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{N}+\mathrm{1}\right)}\right) \\ $$$$=\mathrm{H}_{\mathrm{N}} +\mathrm{ln}\left(\frac{\Gamma\left(\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{N}\right).\mathrm{N}}\right)−\mathrm{2ln}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{H}_{\mathrm{N}} −\mathrm{ln}\left(\mathrm{N}\right)+\mathrm{2ln}\left(\frac{\Gamma\left(\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{N}\right).\sqrt{\mathrm{N}}}\right)−\mathrm{2ln}\left(\sqrt{\pi}\right) \\ $$$$\Omega=\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}}\Omega_{\mathrm{N}} =\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{H}_{\mathrm{N}} −\mathrm{ln}\left(\mathrm{N}\right)−\mathrm{ln}\left(\pi\right)\right) \\ $$$$+\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}log}\left(\frac{\Gamma\left(\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{N}\right)\sqrt{\mathrm{N}}}\right)− \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{H}_{\mathrm{x}} −\mathrm{ln}\left(\mathrm{x}\right)=\gamma \\ $$$$\Gamma\left(\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sim\sqrt{\mathrm{2}\pi}\left(\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\left(\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\Gamma\left(\mathrm{N}\right)\:\sim\sqrt{\mathrm{2}\pi}\mathrm{N}^{\mathrm{N}} \mathrm{e}^{−\mathrm{N}} \\ $$$$\Omega=\gamma−\mathrm{log}\left(\pi\right)+\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}}.\mathrm{2log}\left(\frac{\sqrt{\mathrm{2}\pi}\left(\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\mathrm{N}−\frac{\mathrm{1}}{\mathrm{2}}} }{\:\sqrt{\mathrm{2}\pi}.\mathrm{N}^{\mathrm{N}} \mathrm{e}^{−\mathrm{N}} .\sqrt{\mathrm{N}}}\right) \\ $$$$\gamma−\mathrm{log}\left(\pi\right)+\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}}.\mathrm{2log}\left(\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2N}}\right)^{\mathrm{N}} .\frac{\sqrt{\mathrm{N}+\frac{\mathrm{1}}{\mathrm{2}}}}{\mathrm{N}}.\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$$=\gamma−\mathrm{log}\left(\pi\right) \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}+\mathrm{2log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}}\right)=\gamma−\mathrm{log}\left(\pi\right) \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 31/Dec/23
$$\: \\ $$
Commented by witcher3 last updated on 31/Dec/23
thank You sir have a nice day
$$\mathrm{thank}\:\mathrm{You}\:\mathrm{sir}\:\mathrm{have}\:\mathrm{a}\:\mathrm{nice}\:\mathrm{day} \\ $$

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