Question

Solve: $\underset{n\to \theta }{lim}\left(\frac{1^k+2^k+3^k+.....k}{n^{k+1}}\right)=(k>-1)$

Answer 1

$\underset{n\to \theta }{lim}\left(\frac{1^k+2^k+3^k+.....k}{n^{k+1}}\right)=(k3-1)$

By definition of the Riemann integral we have:

$\underset{n\to \infty }{lim}\frac{1}{n}\sum _{i=1}^n{\left(\frac{i}{n}\right)}^k$

$={\int }_0^1{x}^{k}dx$

$=\frac{1}{1+k}$.