Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 204991 by tigrecomplexe last updated on 04/Mar/24

Commented by tigrecomplexe last updated on 05/Mar/24

$$sommeonecanputhandhereplease?$$

Answered by witcher3 last updated on 05/Mar/24

$${\mathrm{u}}_{\mathrm{n}}={\underset{\mathrm{n}\to \mathrm{\infty }}{\lim e}}^{\sin \left(2\pi \sqrt{1+{\mathrm{n}}^2}\right)\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{\ln (1+\frac{\mathrm{k}}{\mathrm{n}})}{2}}$$$$\sin \left(2\pi \sqrt{1+{\mathrm{n}}^2}\right)=\sin \left(2\pi \mathrm{n}{(1+\frac{1}{{\mathrm{n}}^2})}^{\frac{1}{2}}\right)$$$${(1+\frac{1}{{\mathrm{n}}^2})}^{\frac{1}{2}}=1+\frac{1}{{2\mathrm{n}}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)$$$$\sin \left(2\pi \sqrt{1+{\mathrm{n}}^2}\right)\stackrel{\mathrm{n}\to \mathrm{\infty }}{=}\sin (2\pi \mathrm{n}+\frac{\pi }{\mathrm{n}}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right))=\sin (\frac{\pi }{\mathrm{n}}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right))$$$$\sin \left(2\pi \sqrt{1+{\mathrm{n}}^2}\right)\stackrel{\mathrm{\infty }}{=}\frac{\pi }{\mathrm{n}}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)$$$${\mathrm{u}}_{\mathrm{n}}={\mathrm{e}}^{\mathrm{nsin}\left(2\pi \sqrt{1+{\mathrm{n}}^2}\right).\frac{1}{2\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\ln (1+\frac{\mathrm{k}}{\mathrm{n}})}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }.\frac{1}{2}.\frac{1}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\ln (1+\frac{\mathrm{k}}{\mathrm{n}})={\int }_0^1\frac{\ln (1+\mathrm{x})}{2}={[\frac{(1+\mathrm{x})}{2}\ln (1+\mathrm{x})-\frac{\mathrm{x}}{2}]}_0^1=\ln \left(2\right)-\frac{1}{2}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }.\mathrm{nsin}\left(2\pi \sqrt{1+{\mathrm{n}}^2}\right)=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\mathrm{n}.\left(\frac{\pi }{\mathrm{n}}\right)=\pi$$$${\mathrm{U}}_{\mathrm{n}}\to {\mathrm{e}}^{\pi (\ln \left(2\right)-\frac{1}{2})}=\frac{2\pi }{{\mathrm{e}}^{\frac{\pi }{2}}}$$$$$$

Answered by Mathspace last updated on 05/Mar/24

$$A_n={\left(\prod _{k=1}^n\sqrt{1+\frac{k}{n}}\right)}^{sin\left(2\pi \sqrt{1+n^2}\right)}$$$$\Rightarrow log\left(A_n\right)=sin(2\pi \sqrt{1+n^2)}\sum _{k=1}^{n}log\left(\sqrt{1+\frac{k}{n}}\right)$$$$\sqrt{1+n^2}=n\sqrt{1+\frac{1}{n^2}}=n{(1+\frac{1}{n^2})}^{\frac{1}{2}}$$$$\sim n(1+\frac{1}{2n^2})=n+\frac{1}{2n}\Rightarrow$$$$sin\left(2\pi \sqrt{1+n^2}\right)\sim sin(2\pi n+\frac{\pi }{n})$$$$=sin\left(\frac{\pi }{n}\right)\sim \frac{\pi }{n}\Rightarrow$$$$log\left(A_n\right)\sim \frac{\pi }{n}\sum _{k=1}^{n}log\sqrt{1+\frac{k}{n}}$$$$Reimansumgive$$$${lim}_{n\to +\mathrm{\infty }}log\left(A_n\right)$$$$=\pi {\int }_0^{1}log\sqrt{1+x}dx$$$$=\frac{\pi }{2}{\int }_0^{1}log(1+x)dxbut$$$${\int }_0^{1}log(1+x)dx=$$$${\left[xlog(1+x)\right]}_0^1-{\int }_0^1\frac{x}{1+x}dx$$$$=log2-{\int }_0^1\frac{1+x-1}{1+x}dx$$$$=log2-1+{\int }_0^1\frac{dx}{1+x}$$$$=log2-1+{\left[log(1+x)\right]}_0^1$$$$=log2-1+log2=2log2-1$$$$limlog\left(A_n\right)=\frac{\pi }{2}(2log2-1)$$$$=\pi log2-\frac{\pi }{2}\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{A}_n=e^{\pi log2-\frac{\pi }{2}}$$$$=2^{\pi }\times e^{-\frac{\pi }{2}}$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com