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.