infinite series and Big ''O''

Question

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 $$\sum_{n=M}^{\infty}\frac{1}{n^{2}}=O(\frac{1}{M})$$

Hint: Can you compare the sum with an integral? – Nov 05 '12 at 18:51 — , Nov 05 '12 at 18:51
This bound was also discussed at this MSE link. – Marko Riedel Apr 08 '14 at 22:22 — Marko Riedel, Apr 08 '14 at 22:22

score 4 · Answer 1 · answered Nov 05 '12 at 18:52

4

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

answered Nov 05 '12 at 18:52

infinite series and Big ''O''

1 Answers1