Question

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

$\frac{1}{p+1}$

$\underset{n\to \infty }{lim}\frac{1^n+2^n+...+n^n}{n^p}\times \frac{1}{n}$
$=\underset{n\to \infty }{lim}\frac{1}{n}[{\left(\frac{1}{n}\right)}^p+{\left(\frac{2}{n}\right)}^p+...+{\left(\frac{n}{n}\right)}^p]$
$=\underset{n\to \infty }{lim}\sum _{r=1}^{r=n}\frac{1}{n}{\left(\frac{r}{n}\right)}^p$
$={\int }_0^1{x}^{n}dx=\frac{1}{p+1}$
Ans: A