數列\(a_1, a_2, \dots, a_n, \dots\) 有關係式\(\displaystyle \log_{n+1} a_n = 1 + \frac{1}{(n+1)\log(n+1)}\)。若\(\displaystyle \frac{a_n}{n+1} < 1.6\)，則自然數\(n\)之最小值為 _______。

\(\displaystyle a_n = (n+1)^{\left(1 + \frac{1}{(n+1)\log(n+1)}\right)}\)則
\(\displaystyle \frac{a_n}{n+1} = (n+1)^{\frac{1}{(n+1)\log(n+1)}} < 1.6\)
取對數得\(\displaystyle \frac{1}{(n+1)\log(n+1)} \cdot \log(n+1) < \log(1.6)\)
\(\displaystyle \frac{1}{n+1} < \log(1.6)=0.204\)
\(n\)取4

感謝大家喔！