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We all know that $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$. If $M$ is a positive integer, how can we show that $$\sum_{n=M}^{\infty}\frac{1}{n^{2}}=O(\frac{1}{M})$$

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Hint: Can you compare the sum with an integral? – user27126 Nov 5 '12 at 18:51
@Sanchez: Good hint, and +1 for giving it as a hint. – Did Nov 5 '12 at 18:53
This bound was also discussed at this MSE link. – Marko Riedel Apr 8 '14 at 22:22

1 Answer 1

$$\sum_{n=M}^{\infty} \dfrac1{n^2}< \int_{M-1}^{\infty} \dfrac{dx}{x^2} = -\left . \dfrac1x \right \vert_{M-1}^{\infty} = \dfrac1{M-1} = \mathcal{O} \left(\dfrac1M\right)$$

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thank you so much.. – John Nov 5 '12 at 19:21

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