$\lim\limits_{n \rightarrow \infty}\left\{\frac{1^m+2^m+3^m+...+n^m}{n^{m+1}}\right\}$ equals
Answer & explanation
Correct answer: option 1
We have,
$\lim\limits_{n \rightarrow \infty}\left\{\frac{1^m+2^m+...+n^m}{n^{m+1}}\right\}$
$=\lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n \frac{r^m}{n^m} . \frac{1}{n}=\lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n\left(\frac{r}{n}\right)^m \frac{1}{n}=\int\limits_0^1 x^m d x=\frac{1}{m+1}$