Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 155604 by SANOGO last updated on 02/Oct/21

Answered by Kamel last updated on 02/Oct/21

$$$$$$\mathrm{\forall }x⩾0x-\frac{x^3}{6}⩽sin\left(x\right)⩽x$$$$\therefore \underset{k=1}{\sum ^n}\frac{1}{n+k}-\frac{1}{6}\underset{k=1}{\sum ^n}\frac{1}{{(n+k)}^3}⩽\underset{k=1}{\sum ^n}sin\left(\frac{1}{n+k}\right)⩽\underset{k=1}{\sum ^n}\frac{1}{n+k}$$$$\therefore \underset{n\to +\mathrm{\infty }}{lim}\frac{1}{n}\underset{k=1}{\sum ^n}\frac{1}{1+\frac{k}{n}}-\frac{1}{6}\underset{n\to +\mathrm{\infty }}{lim}\frac{1}{n^2}\frac{1}{n}\underset{k=1}{\sum ^n}\frac{1}{{(1+\frac{k}{n})}^3}⩽\underset{n\to +\mathrm{\infty }}{lim}\underset{k=1}{\sum ^n}sin\left(\frac{1}{n+k}\right)⩽\underset{n\to +\mathrm{\infty }}{lim}\frac{1}{n}\underset{k=1}{\sum ^n}\frac{1}{1+\frac{k}{n}}$$$$So:{\int }_0^1\frac{dx}{1+x}-\underset{n\to +\mathrm{\infty }}{lim}\frac{1}{n^2}{\int }_0^1\frac{dx}{{(1+x)}^3}⩽\underset{n\to +\mathrm{\infty }}{lim}\underset{k=1}{\sum ^n}sin\left(\frac{1}{n+k}\right)⩽{\int }_0^1\frac{dx}{1+x}$$$$Then:\underset{n\to +\mathrm{\infty }}{lim}\underset{k=1}{\sum ^n}sin\left(\frac{1}{n+k}\right)=Ln\left(2\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com