Question

$\underset{n\to \infty }{\lim }\frac{1^p+2^p+3^p+....+n^p}{n^{p+1}}$ is

• 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

Answer: (A)

We have $\underset{n\to \infty }{\lim }\frac{1^p+2^p+....+n^p}{n^{p+1}};$
$\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{r^p}{n^p.n}=\underset{0}{\overset{1}{\int }}{x}^{p}dx={\left[\frac{x^{p+1}}{p+1}\right]}_0^1=\frac{1}{p+1}$