$\bf命题1：$设$f(x)$是$\left[ {1, + \infty } \right)$上的非负单调减少函数，令
a
n=∑k=1nf(k)?∫n1f(x)dx,n∈N+
证明：数列$\left\{ {{a_n}} \right\}$收敛
证明：由$f(x)$在$\left[ {1, + \infty } \right)$上单调减少知，$f(x)$在$\left[ {n,n + 1}
\right]$上可积，且
f(n+1)≤∫n+1nf(x)dx≤f(n),n∈N+
从而可知
an+1?an=f(n+1)?∫n+1nf(x)dx≤0
即$\left\{ {{a_n}} \right\}$单调减少；而又由$\eqref {eq1}$知
an=∑k=1nf(k)?∫n1f(x)dx≥∑k=1n∫k+1kf(x)dx?∫n1f(x)dx=∫n+1nf(x)dx≥f(n+1)≥0
即$\left\{ {{a_n}} \right\}$有上界，故由$\eqref {eq2}$,$\eqref
{eq3}$及单调有界原理知数列$\left\{ {{a_n}} \right\}$收敛5656,布布扣,bubuko.com
5656
原文：http://www.cnblogs.com/ly758241/p/3706437.html
