回復 27# tsusy 的帖子
應該是這樣吧?
證:\(\displaystyle \frac{log n}{log(n-1)}>\frac{log(n+1)}{log n}\)
pf:
\(\displaystyle \frac{log n^2}{2}>\frac{log(n-1)+log(n+1)}{2}\ge \sqrt{log(n-1)log(n+1)}\)
\(\displaystyle log n>\sqrt{log(n-1)log(n+1)}\)
\((log n)^2>log(n-1)log(n+1)\)