3楼: Originally posted by
Edstrayer at 2016-03-30 12:14:15
\lim\limits_{n\to+\infty}\left(\frac{1}{\frac{1}{n}-\frac{1}{2n^2}}\ln\left(\sum\limits_{k=0}^n\frac{1}{k!}\right)-n\right)
=\lim\limits_{n\to+\infty}\frac{n}{2n-1}\left(2n\ln\left(\sum\limits_{k= ...