fï¼xï¼=x-1-lnxï¼f '(x)=1-1/xï¼å½x>1æ¶ï¼f '(x)>0ï¼f (x)åè°éå¢ã
æ以å½x>1æ¶ï¼f(x)>f(1)=0ã
åf(2)=1-ln2>0
f(3/2)=1/2-ln(3/2)>0
f(4/3)=1/3-ln(4/3)>0
......
f[(n+1)/n]=1/n-ln[(n+1)/n]>0
ç´¯å å¾ï¼1+1/2+1/3+â¦â¦+1/n>ln2+ln(3/2)+ln(4/3)+â¦â¦+ln[(n+1)/n]=ln(n+1)
åln(n+1)-1/2*ln((n+1)(n+2)/2)=1/2ln[(n+1)^2/((n+1)(n+2)/2)]=1/2ln[(2n+2)/(n+2)]>0
å æ¤åä¸çå¼æç«ã
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