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 187983 by TUN last updated on 24/Feb/23

$$\underset{x\to \mathrm{\infty }}{\lim }(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{2x})$$

Answered by cortano12 last updated on 24/Feb/23

$$=\underset{x\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\left(\frac{1}{\mathrm{x}+\mathrm{k}}\right)$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{x}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\left(\frac{1}{1+\frac{\mathrm{k}}{\mathrm{x}}}\right)$$$$=\underset{0}{\stackrel{1}{\int }}\left(\frac{1}{1+\mathrm{x}}\right)\mathrm{dx}=\ln 2$$

Commented by Ar Brandon last updated on 24/Feb/23

$$\underset{k=1}{\sum ^x}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com