Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 123813 by snipers237 last updated on 28/Nov/20

$$\underset{x\to 0}{\lim }{}^x\sqrt{\frac{1}{n}\underset{k=1}{\sum ^n}{k}^x}={}^n\sqrt{n!}$$

Answered by Dwaipayan Shikari last updated on 28/Nov/20

$$\left(\frac{1}{n}(1^x+2^x+3^x+...+n^x)\right)⩾{(1^x.2^x.3^x.4^x..n^x)}^{\frac{1}{n}}$$$$\frac{1}{n}\sum ^n{k}^x⩾{(n!)}^{\frac{x}{n}}$$$${\left(\frac{1}{n}\sum ^n{k}^x\right)}^{\frac{1}{x}}⩾{(n!)}^{\frac{1}{n}}$$$$Asx\to 0{\left(\frac{1}{n}\sum ^n{k}^x\right)}^{\frac{1}{x}}={(n!)}^{\frac{1}{n}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com