# 积分判别法

（重定向自積分檢驗

## 证明

${\displaystyle \int _{n}^{n+1}f(x)\,dx}$

${\displaystyle f(n+1)\leq \int _{n}^{n+1}f(x)\,dx\leq f(n)}$

${\displaystyle \sum _{n=1}^{k}f(n+1)\leq \sum _{n=1}^{k}\int _{n}^{n+1}f(x)\,dx\leq \sum _{n=1}^{k}f(n)}$

${\displaystyle \sum _{n=1}^{k}\int _{n}^{n+1}f(x)\,dx=\int _{1}^{k+1}f(x)\,dx}$

${\displaystyle \sum _{n=1}^{\infty }f(n+1)\leq \int _{1}^{\infty }f(x)\,dx\leq \sum _{n=1}^{\infty }f(n)}$

## 例子

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}}$

${\displaystyle \int _{1}^{M}{\frac {1}{x}}\,dx=\ln x{\Bigr |}_{1}^{M}=\ln M\to \infty }$ ，当${\displaystyle M\to \infty }$ 时。

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{1+\varepsilon }}}}$

${\displaystyle \int _{1}^{M}{\frac {1}{x^{1+\varepsilon }}}\,dx=-{\frac {1}{\varepsilon x^{\varepsilon }}}{\biggr |}_{1}^{M}={\frac {1}{\varepsilon }}{\Bigl (}1-{\frac {1}{M^{\varepsilon }}}{\Bigr )}\leq {\frac {1}{\varepsilon }}}$ ，对于所有${\displaystyle M\geq 1.}$

## 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