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 166572 by qaz last updated on 22/Feb/22

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}{2}^{-\mathrm{k}}\cos \sqrt{\frac{\mathrm{k}}{\mathrm{n}}}=?$$

Answered by alephzero last updated on 23/Feb/22

$$\underset{n\to \mathrm{\infty }}{\lim }\underset{k=0}{\sum ^{2n}}{2}^{-k}\cos \sqrt{\frac{k}{n}}=?$$$$\mathrm{If}n\to \mathrm{\infty }\Rightarrow \frac{k}{n}\to 0\wedge 2n\to \mathrm{\infty }$$$$\Rightarrow \underset{k=0}{\sum ^{\mathrm{\infty }}}{2}^{-k}\cos \sqrt{\frac{k}{n}}=$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}{2}^{-k}\cos \sqrt{0}=\underset{k=0}{\sum ^{\mathrm{\infty }}}{2}^{-k}=$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2^k}=1+\xi \left(2\right)=1+\frac{1}{2-1}=$$$$=1+1=2$$$$\xi \left(s\right)=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{s^k}\mathrm{is}\mathrm{my}\mathrm{function}.$$$$\mathrm{One}\mathrm{day}\mathrm{I}\mathrm{was}\mathrm{studying}\mathrm{convergness}$$$$\mathrm{of}\mathrm{this}\mathrm{series}.\mathrm{I}\mathrm{noticed},\mathrm{that}$$$$\xi \left(s\right)=\frac{1}{s-1}$$$$\mathrm{I}\mathrm{cheked}\mathrm{this}\mathrm{with}\mathrm{Wolphram}$$$$\mathrm{Alpha}.\mathrm{It}\mathrm{was}\mathrm{right}.\mathrm{So},\mathrm{this}$$$$\mathrm{function}\mathrm{reveals}\mathrm{that}\frac{0}{0}=-1$$$$\left(\mathrm{this}\mathrm{is}\mathrm{weird}\mathrm{indeed}\right)$$

Commented by MJS_new last updated on 22/Feb/22

$$\frac{0}{0}\mathrm{is}\mathrm{not}\mathrm{defined}\Rightarrow \frac{0}{0}=-1\mathrm{is}\mathrm{wrong}$$$$\mathrm{but}\mathrm{you}\mathrm{can}\mathrm{find}\mathrm{limits}\mathrm{of}\mathrm{the}\mathrm{form}$$$$L=\underset{x\to r}{\lim }\frac{f\left(x\right)}{g\left(x\right)}\mathrm{with}f\left(r\right)=g\left(r\right)=0\mathrm{and}\mathrm{any}$$$$\mathrm{real}\mathrm{value}\mathrm{of}L.\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{proof}\mathrm{for}$$$$\frac{0}{0}=L\in \mathbb{R}(\mathrm{because}L\mathrm{can}\mathrm{be}\mathrm{any}\mathrm{number}\mathrm{we}$$$$\mathrm{would}\mathrm{get}\mathrm{i}.\mathrm{e}.\frac{0}{0}=-1=1={\pi }^2=\sqrt{37}\mathrm{e}...\times )$$

Answered by Mathspace last updated on 22/Feb/22

$$cosu=1-\frac{u^2}{2}+o\left(u^2\right)\Rightarrow$$$$1-\frac{u^2}{2}⩽cosu⩽1\Rightarrow$$$$1-\frac{{\left(\sqrt{\frac{k}{n}}\right)}^2}{2}⩽cos\sqrt{\frac{k}{n}}⩽1\Rightarrow$$$$2^{-k}(1-\frac{k}{2n})⩽2^{-k}cos\sqrt{\frac{k}{n}}⩽2^{-k}\Rightarrow$$$$\frac{1}{2^k}-\frac{1}{2n}\frac{k}{2^k}⩽2^{-k}cos\sqrt{\frac{k}{n}}⩽\frac{1}{2^k}\Rightarrow$$$$\Rightarrow \sum _{k=0}^{2n}\frac{1}{2^k}-\frac{1}{2n}\sum _{k=0}^{2n}\frac{k}{2^k}⩽\sum _{k=0}^{2n}{2}^{-k}cos\sqrt{\frac{k}{n}}⩽\sum _{k=0}^{2n}\frac{1}{2^k}$$$$wehave\sum _{k=0}^{2n}{\left(\frac{1}{2}\right)}^k=\frac{1-{\left(\frac{1}{2}\right)}^{2n+1}}{1-\frac{1}{2}}$$$$=2(1-\frac{1}{2^{2n+1}})\to 2$$$$\sum _{k=0}^N{x}^k=\frac{1-x^{N+1}}{1-x}\Rightarrow$$$$\sum _{k=1}^N{kx}^{k-1}=\frac{d}{dx}\left(\frac{x^{N+1}-1}{x-1}\right)$$$$=\frac{{Nx}^{N+1}-(N+1)x^N+1}{{(1-x)}^2}\Rightarrow$$$$\Rightarrow \sum _{k=1}^{2n}{kx}^k=\frac{x}{{(1-x)}^2}(2{nx}^{2n+1}-(2n+1)x^{2n}+1)\Rightarrow$$$$\sum _{k=1}^{2n}\frac{k}{2^k}=\frac{1}{2{(1-\frac{1}{2})}^2}(\frac{2n}{2^{2n+1}}-\frac{2n+1}{2^{2n}}+1)$$$$=2(...)\to 2\Rightarrow \frac{1}{2n}\sum _{k=1}^{2n}\frac{k}{2^k}\to 0\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}\sum _{k=0}^{2n}{2}^{-k}cos\sqrt{\frac{k}{n}}=2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com