回復 1# eyeready 的帖子
\(\begin{align}
& \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{\left[ \left( n+1 \right)\left( n+2 \right)\cdots \left( n+n \right) \right]}^{\frac{1}{n}}}}{n} \\
& =\underset{n\to \infty }{\mathop{\lim }}\,{{\left[ \left( 1+\frac{1}{n} \right)\left( 1+\frac{2}{n} \right)\cdots \left( 1+\frac{n}{n} \right) \right]}^{\frac{1}{n}}} \\
& ={{e}^{\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\left[ \ln \left( 1+\frac{1}{n} \right)+\ln \left( 1+\frac{2}{n} \right)+\cdots +\ln \left( 1+\frac{n}{n} \right) \right]}} \\
& ={{e}^{\int_{1}^{2}{\ln xdx}}} \\
& ={{e}^{\ln 4-1}} \\
& =\frac{4}{e} \\
\end{align}\)
