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 122458 by benjo_mathlover last updated on 17/Nov/20

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{\sqrt{n}}(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}})=?$$

Answered by liberty last updated on 17/Nov/20

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

Answered by Dwaipayan Shikari last updated on 17/Nov/20

$$\frac{1}{2\sqrt{n}}(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+....+\sqrt{2n+1}-\sqrt{2n-1})$$$$\frac{1}{2\sqrt{n}}(\sqrt{2n+1}-1)=\frac{1}{2}(\sqrt{2+\frac{1}{n}}-\frac{1}{2\sqrt{n}})=\frac{1}{\sqrt{2}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com