积分判别法

$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{k^s}$

证明

$\int_n^{n+1} f(x)\, dx$

$f(n+1) \leq \int_n^{n+1} f(x)\, dx \leq f(n)$

$\sum_{n=1}^{k} f(n+1) \leq \sum_{n=1}^{k} \int_n^{n+1} f(x)\, dx \leq \sum_{n=1}^{k} f(n)$

$\sum_{n=1}^{k} \int_n^{n+1} f(x)\, dx= \int_1^{k+1} f(x)\, dx$

$\sum_{n=1}^{\infty} f(n+1) \leq \int_1^{\infty} f(x)\, dx \leq \sum_{n=1}^{\infty} f(n)$

例子

$\sum_{n=1}^\infty \frac1n$

$\int_1^M\frac1x\,dx=\ln x\Bigr|_1^M=\ln M\to\infty$，当$M\to\infty$时。

$\sum_{n=1}^\infty \frac1{n^{1+\varepsilon}}$

$\int_1^M\frac1{x^{1+\varepsilon}}\,dx =-\frac1{\varepsilon x^\varepsilon}\biggr|_1^M= \frac1\varepsilon\Bigl(1-\frac1{M^\varepsilon}\Bigr) \le\frac1\varepsilon$，对于所有$M\ge1.$

參考

• Knopp, Konrad, "Infinite Sequences and Series", Dover publications, Inc., New York, 1956. (§ 3.3) ISBN 0486601536
• Whittaker, E. T., and Watson, G. N., A Course in Modern Analysis, fourth edition, Cambridge University Press, 1963. (§ 4.43) ISBN 0521588073