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.