Asymptotic relation of $\sum_{n > x}\frac{1}{n^2}$

Asked Dec 09 '20 at 08:17

Active Jun 20 '22 at 16:33

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In Apostol's Analytic number theory I encountered the following asymptotic relation. I do not know how this is obtained. Since, I am a beginner to this subject kindly provide a detailed explanation and point me to some resources where I can study more about this.

$\displaystyle O\left (\sum_{n > x}\frac{1}{n^2}\right) = O\left (\frac{1}{x} \right)$

edited Jun 20 '22 at 16:33

asked Dec 09 '20 at 08:17

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Just use bound by $\int_x^{\infty} \frac 1 {t^{2}} dt $ as you do in the Integral test for convergence of a series of positive terms. – Kavi Rama Murthy Dec 09 '20 at 08:20
Hint: approximate the sum by an integral $\sum\limits_{n>x}\frac{1}{n^2} \sim \int_x^\infty \frac{dt}{t^2} = \frac1x$ – achille hui Dec 09 '20 at 08:21
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Or bound it using $\sum\limits_{n\geqslant x}\frac{1}{n(n+1)}$ evaluated exactly. – metamorphy Dec 12 '20 at 10:45

Asymptotic relation of $\sum_{n > x}\frac{1}{n^2}$

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