问题及解答

证明:

\[
\sqrt[n]{n}-1\sim\frac{\ln n}{n},\quad\text{when}\ n\rightarrow+\infty
\]

\[
\sqrt[n]{n}-1=n^{\frac{1}{n}}-1=e^{\frac{1}{n}\ln n}-1\quad\sim\frac{\ln n}{n}
\]

这里应用了等价无穷小: 

\[
e^x-1\sim x,\quad\text{when}\ x\rightarrow 0.
\]