Question

${lim}_{n\to \infty }$$\frac{l^m+2^m+3^m+...+n^m}{n^{m+1}}$

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

Answer 1

Answer: (A)

$\frac{1}{m+1}$

$\underset{n\to \infty }{lim}\frac{1^m+2^m+....+n^m}{n^{m+1}}=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{r^m}{n^{m+1}}$

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}{\left(\frac{r}{n}\right)}^m={\int }_0^1{x}^{m}dx=\frac{1}{m+1}$