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 $\sum_{n=M}^{\infty}\frac{1}{n^{2}}=O(\frac{1}{M})$
infinite series and Big ''O''
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calculus
real-analysis
sequences-and-series
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0This bound was also discussed at this [MSE link](http://math.stackexchange.com/questions/685435/trying-to-get-a-bound-on-the-tail-of-the-series-for-zeta2). – 2014-04-08
1 Answers
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$\sum_{n=M}^{\infty} \dfrac1{n^2}< \int_{M-1}^{\infty} \dfrac{dx}{x^2} = -\left . \dfrac1x \right \vert_{M-1}^{\infty} = \dfrac1{M-1} = \mathcal{O} \left(\dfrac1M\right)$