Let $\displaystyle N > M $. Prove that for $\displaystyle k > 1 $, $\displaystyle \sum_{n=M+1}^{N} n^{-k} = O(M^{-k+1}) $ as $\displaystyle M \to \infty $.

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- Sep 10th 2009, 07:53 AMmathman88Big O
Let $\displaystyle N > M $. Prove that for $\displaystyle k > 1 $, $\displaystyle \sum_{n=M+1}^{N} n^{-k} = O(M^{-k+1}) $ as $\displaystyle M \to \infty $.

- Sep 11th 2009, 06:52 AMWalter Von Mondale
Observe your sum is bounded above by the integral of 1 \x^k from M to N-1, which you can easily evaluate directly obtaining two terms bounded by C/M^(k-1) as needed.