Question #87417

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Answer 1 · 2017-07-22T13:25:46

See bellow.

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.