解由任意的x1,x2∈(-∞,0］,有(x1-x2)［f(x2)-f(x1)］＞0知当x1-x2＞0时,f(x2)-f(x1)＞0,即x1＞x2时,f(x1)＜f(x2)知f(x)在(-∞,0］是减函数,又由f(x)是偶函数故f(x)在（0,正无穷大）是增函数又由0≤/n-1/＜/n/＜/n+1...那么说来是f(-n)＜f(n-1)＜f(n＋1)吗？