$f命题1：$设$f(x)$是$left[ {1, + infty } ight)$上的非负单调减少函数，令
[{a_n} = sumlimits_{k = 1}^n {fleft( k
ight)} - int_1^n {fleft( x
ight)dx} ,n in {N_ + }]
证明：数列$left{ {{a_n}}
ight}$收敛
证明：由$f(x)$在$left[ {1, + infty }
ight)$上单调减少知，$f(x)$在$left[ {n,n + 1}
ight]$上可积，且
egin{equation}label{eq1}fleft( {n + 1}
ight) le int_n^{n + 1} {fleft( x
ight)dx} le fleft( n
ight),n in {N_ + }end{equation}
从而可知egin{equation}label{eq2}{a_{n + 1}} - {a_n} = fleft( {n + 1}
ight) - int_n^{n + 1} {fleft( x
ight)dx} le 0end{equation}
即$left{ {{a_n}}
ight}$单调减少；而又由$eqref {eq1}$知
egin{align}label{eq3}
{a_n} &= sumlimits_{k = 1}^n {fleft( k
ight)} - int_1^n {fleft( x
ight)dx}
onumber\&
ge sumlimits_{k = 1}^n {int_k^{k + 1} {fleft( x
ight)dx} } - int_1^n {fleft( x
ight)dx}\&
= int_n^{n + 1} {fleft( x
ight)dx} ge fleft( {n + 1}
ight) ge 0
onumberend{align}
即$left{ {{a_n}}
ight}$有上界，故由$eqref {eq2}$,$eqref {eq3}$及单调有界原理知数列$left{ {{a_n}}
ight}$收敛