f(x)=lg (1-x)/(1+x) 若n为整数,求证,f(1/5）+f(1/11)+..+f(1/n*n+3n+1)>f(1/2)

题目

f(x)=lg (1-x)/(1+x) 若n为整数,求证,f(1/5）+f(1/11)+..+f(1/n*n+3n+1)>f(1/2)

答案

f(1/n*n+3n+1) = lg (1-(1/n*n+3n+1) )/(1+（1/n*n+3n+1))
= lg[ (n*n+3n) / (n*n+3n+2)]
=lg[n(n+3)/ ((n+1)*(n+2))]
=lgn + lg(n+3) - lg(n+1)-lg(n+2)
所以f(1/5）+f(1/11)+..+f(1/n*n+3n+1)
= lg1 + lg(1+3) - lg(1+1)-lg(1+2) +lg2 + lg(2+3) - lg(2+1)-lg(2+2)
+lg3 + lg(3+3) - lg(3+1)-lg(3+2) +lg4 + lg(4+3) - lg(4+1)-lg(4+2)
+.+lgn + lg(n+3) - lg(n+1)-lg(n+2)
=lg(n+3) - lg(n+1) - lg3
=lg[((n+3) /(3n+3) ]
>lg(1/3) = f(1/2)