Question

$\underset{n\to \infty }{lim}\frac{1^p+2^p+3^p+......+n^p}{n^{p+1}}$ is equal to (p$\ne$ -1)

• A. $\frac{1}{p+1}$
• B. $\frac{1}{1-p}$
• C. $\frac{1}{p}-\frac{1}{p-1}$
• D. $\frac{1}{p+2}$

Answer 1

The correct option is A $\frac{1}{p+1}$

$l=\underset{x\to \infty }{lim}\frac{1^p+2^p+3^p+{....n}^p}{n^{p+1}}=\underset{x\to \infty }{lim}\frac{1^p+2^p+3^p+{....n}^p}{{n.n}^p}=\frac{1}{n}\underset{x\to \infty }{lim}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^p$

Replacing, $\frac{r}{n}\to x\frac{1}{n}\to dx\sum \to \int$

We have, $l={\int }_0^1{x}^{p}dx=[\frac{x^{p+1}}{p+1}\stackrel{1}{\underset{0}{]}}=\frac{1}{p+1}$