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 192274 by York12 last updated on 13/May/23

$$provethat\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{k=1}{\sum ^n}\frac{n^2}{\sqrt{n^6+k}}\right)=1$$$$$$$$$$

Answered by aleks041103 last updated on 14/May/23

$$\underset{k=1}{\sum ^n}\frac{n^2}{\sqrt{n^6+n}}<\underset{k=1}{\sum ^n}\frac{n^2}{\sqrt{n^6+k}}<\underset{k=1}{\sum ^n}\frac{n^2}{\sqrt{n^6+0}}$$$$\frac{n^3}{n^3\sqrt{1+n^{-5}}}<\underset{k=1}{\sum ^n}\frac{n^2}{\sqrt{n^6+k}}<\frac{n^3}{n^3}$$$$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n^5}}}<\underset{k=1}{\sum ^n}\frac{n^2}{\sqrt{n^6+k}}<1$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{1+\frac{1}{n^5}}}⩽\underset{n\to \mathrm{\infty }}{lim}\left(\underset{k=1}{\sum ^n}\frac{n^2}{\sqrt{n^6+k}}\right)⩽1$$$$\Rightarrow 1⩽\underset{n\to \mathrm{\infty }}{lim}\left(\underset{k=1}{\sum ^n}\frac{n^2}{\sqrt{n^6+k}}\right)⩽1$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\left(\underset{k=1}{\sum ^n}\frac{n^2}{\sqrt{n^6+k}}\right)=1$$

Commented by York12 last updated on 14/May/23

$$actullyIamahighschoolstudentingrade$$$$10andIwaswonderingwherecanIlearnthat$$$$Ihopeifyoucanhelpmeoutwiththat$$

Commented by York12 last updated on 14/May/23

$$sointrestingtechnique$$

Commented by York12 last updated on 14/May/23

$$mytelegram:yorkgubler$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com