# Big O

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• September 10th 2009, 08:53 AM
mathman88
Big O
Let $N > M$. Prove that for $k > 1$, $\sum_{n=M+1}^{N} n^{-k} = O(M^{-k+1})$ as $M \to \infty$.
• September 11th 2009, 07:52 AM
Walter 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.