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Question Number 202589 by mnjuly1970 last updated on 30/Dec/23

$$$$$$\mathrm{\Omega }=\underset{k=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{k}+2ln(1-\frac{1}{2k}))=?$$$$$$$$$$

Commented by mnjuly1970 last updated on 30/Dec/23

$$Yessiryouareright\cancel{\underset{―}{\underbrace{\mathfrak{Y}}}}$$

Commented by MrGHK last updated on 30/Dec/23

$$\mathbf{sir},\mathbf{dont}\mathbf{you}\mathbf{think}\mathbf{it}\mathbf{diverves}?$$

Answered by witcher3 last updated on 30/Dec/23

$$\frac{1}{\mathrm{k}}+2\ln (1-\frac{1}{2\mathrm{k}})\sim \frac{1}{\mathrm{k}}+2(-\frac{1}{2\mathrm{k}}+\frac{1}{{4\mathrm{k}}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{k}}^2}\right))$$$$\sim \frac{1}{{2\mathrm{k}}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{k}}^2}\right)$$$${\mathrm{U}}_{\mathrm{n}}=\frac{1}{{2\mathrm{n}}^2}\mathrm{\Sigma }{\mathrm{U}}_{\mathrm{n}}\mathrm{cv}\Rightarrow \mathrm{\Omega }\mathrm{existe}$$$$\mathrm{\Omega }=\underset{\mathrm{N}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\sum ^{\mathrm{N}}}\frac{1}{\mathrm{k}}+2\ln \left(\frac{2\mathrm{k}-1}{2\mathrm{k}}\right))$$$${\mathrm{\Omega }}_{\mathrm{N}}={\mathrm{H}}_{\mathrm{N}}+2\ln \left(\underset{\mathrm{k}=1}{\prod ^{\mathrm{N}}}\left(\frac{2\mathrm{k}-1}{2\mathrm{k}}\right)\right)={\mathrm{H}}_{\mathrm{N}}+2\ln \left(\frac{\underset{\mathrm{k}=1}{\prod ^{\mathrm{N}}}(\mathrm{k}-\frac{1}{2})}{\underset{\mathrm{k}=1}{\prod ^{\mathrm{N}}}\left(\mathrm{k}\right)}\right)$$$$={\mathrm{H}}_{\mathrm{N}}+2\ln \left(\frac{\mathrm{\Gamma }(\mathrm{N}+\frac{1}{2})}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(\mathrm{N}+1)}\right)$$$$={\mathrm{H}}_{\mathrm{N}}+\ln \left(\frac{\mathrm{\Gamma }(\mathrm{N}+\frac{1}{2})}{\mathrm{\Gamma }\left(\mathrm{N}\right).\mathrm{N}}\right)-2\ln \left(\mathrm{\Gamma }\left(\frac{1}{2}\right)\right)$$$$={\mathrm{H}}_{\mathrm{N}}-\ln \left(\mathrm{N}\right)+2\ln \left(\frac{\mathrm{\Gamma }(\mathrm{N}+\frac{1}{2})}{\mathrm{\Gamma }\left(\mathrm{N}\right).\sqrt{\mathrm{N}}}\right)-2\ln \left(\sqrt{\pi }\right)$$$$\mathrm{\Omega }=\underset{\mathrm{N}\to \mathrm{\infty }}{\lim }{\mathrm{\Omega }}_{\mathrm{N}}=\underset{\mathrm{N}\to \mathrm{\infty }}{\lim }({\mathrm{H}}_{\mathrm{N}}-\ln \left(\mathrm{N}\right)-\ln \left(\pi \right))$$$$+\underset{\mathrm{N}\to \mathrm{\infty }}{\lim log}\left(\frac{\mathrm{\Gamma }(\mathrm{N}+\frac{1}{2})}{\mathrm{\Gamma }\left(\mathrm{N}\right)\sqrt{\mathrm{N}}}\right)-$$$$\underset{x\to 0}{\lim }{\mathrm{H}}_{\mathrm{x}}-\ln \left(\mathrm{x}\right)=\gamma$$$$\mathrm{\Gamma }(\mathrm{N}+\frac{1}{2})\sim \sqrt{2\pi }{(\mathrm{N}+\frac{1}{2})}^{\mathrm{N}+\frac{1}{2}}{\mathrm{e}}^{-(\mathrm{N}+\frac{1}{2})}$$$$\mathrm{\Gamma }\left(\mathrm{N}\right)\sim \sqrt{2\pi }{\mathrm{N}}^{\mathrm{N}}{\mathrm{e}}^{-\mathrm{N}}$$$$\mathrm{\Omega }=\gamma -\log \left(\pi \right)+\underset{\mathrm{N}\to \mathrm{\infty }}{\lim }.2\log \left(\frac{\sqrt{2\pi }{(\mathrm{N}+\frac{1}{2})}^{\mathrm{N}+\frac{1}{2}}{\mathrm{e}}^{-\mathrm{N}-\frac{1}{2}}}{\sqrt{2\pi }.{\mathrm{N}}^{\mathrm{N}}{\mathrm{e}}^{-\mathrm{N}}.\sqrt{\mathrm{N}}}\right)$$$$\gamma -\log \left(\pi \right)+\underset{\mathrm{N}\to \mathrm{\infty }}{\lim }.2\log ({(1+\frac{1}{2\mathrm{N}})}^{\mathrm{N}}.\frac{\sqrt{\mathrm{N}+\frac{1}{2}}}{\mathrm{N}}.{\mathrm{e}}^{-\frac{1}{2}})$$$$=\gamma -\log \left(\pi \right)$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{k}}+2\log (1-\frac{1}{2\mathrm{k}})=\gamma -\log \left(\pi \right)$$$$$$$$$$

Commented by mnjuly1970 last updated on 31/Dec/23

Commented by witcher3 last updated on 31/Dec/23

$$\mathrm{thank}\mathrm{You}\mathrm{sir}\mathrm{have}\mathrm{a}\mathrm{nice}\mathrm{day}$$

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