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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4Hint: Can you compare the sum with an integral? – 2012-11-05
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0@Sanchez: Good hint, and +1 for giving it as a hint. – 2012-11-05
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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