$\bf命题1：$设$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，若$\lim \limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}} f\left( x \right)$存在，则$\lim \limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}} f\left( x \right) = 0$
证明：反证法，若$\lim \limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}} f\left( x \right) = l \ne 0$，则不妨设$l > 0$
从而 由极限的保号性知，存在$M > 0$，当$x>M$时，有\[f\left( x \right) > \frac{1}{2}l\] 于是\[\mathop {\lim }\limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}} \int_M^x {f\left( t \right)dt} \ge \mathop {\lim }\limits_{x \to \begin{array}{*{20}{c}} {{\rm{ + }}\infty } \end{array}} \frac{1}{2}l\left( {x - M} \right) = + \infty \] 这与$\int_a^{ + \infty } {f\left( x \right)dx} $收敛矛盾，故$l=0$
$\bf{注1:}$由于$\int_a^{ + \infty } {f\left( x \right)dx} $收敛，所以$\lim \limits_{A \to \begin{array}{*{20}{c}} { + \infty } \end{array}} \int_a^A {f\left( x \right)dx} $存在
$\bf{注2:}$若$\sum\limits_{n = 1}^\infty {{a_n}} $收敛，则$\lim \limits_{n \to \infty } {a_n} = 0$
原文地址：http://www.cnblogs.com/ly758241/p/3706473.html
