Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
4  
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 at 22:22

1 Answer 1

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

share|improve this answer
    
thank you so much.. –  John Nov 5 '12 at 19:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.