Question

Question #87417

Answer 1

See bellow.

Explanation:

Calling `S_n = sum_(nu=1)^(nu=n)nu^k` we have

`S_(n-1) < n^(k+1)/(k+1) < S_n` or

`0 < n^(k+1)/(k+1)-S_(n-1) < n^k` or

`0 < n/(k+1) - sum_(nu=1)^(nu=n-1)(nu/n)^k < 1`

but

`n sum_(nu=1)^(nu=n-1)(nu/n)^k1/n < n int_0^1 xi^k d xi = n/(k+1)` or

`n sum_(nu=1)^(nu=n-1)(nu/n)^k1/n =n/(k+1)-delta^2` so substituting

`0 < delta^2 < 1`

NOTE:

`sum_(nu=1)^(nu=n-1)(nu/n)^k1/n < int_0^1 xi^k d xi`

because `xi^k` is a monotonically increasing function.