Let $a_n=(1+\frac{1}{n})^n$, show that $$\lim_{n\to\infty}n(e-a_n)\geq \frac e2.$$
We know easily by calculus that $\lim_{n\to\infty}n(e-a_n)=\frac e2$. However, can we give an easier proof without using functional limit, but only the following prosition.
if $a_n>b_n$ for large $n$, then $\lim a_n\geq \lim b_n$.
What I should is just consider $n(a_{n^2}-a_n)$, and show it is $\geq \frac e2$. But is seems not easy to compare $a_{n^2}$ and $a_n$.
We have \begin{align*} \left( {1 + \frac{1}{n}} \right)^n & = \exp \left( {n\log \left( {1 + \frac{1}{n}} \right)} \right) = \exp \left( {1 - \frac{1}{{2n}} + \frac{1}{{3n^2 }} - \cdots } \right) \\ & > \exp \left( {1 - \frac{1}{{2n}}} \right) = e\exp \left( - \frac{1}{{2n}} \right) > e\left( {1 - \frac{1}{{2n}}} \right), \end{align*} i.e., $$ n\left( {e - \left( {1 + \frac{1}{n}} \right)^n } \right) < \frac{e}{2}. $$ Thus, $$ \mathop {\lim }\limits_{n \to + \infty } n\left( {e - \left( {1 + \frac{1}{n}} \right)^n } \right) \le \frac{e}{2}, $$ i.e., the inequality is in the reverse order. Since, in fact, this is an equality, your statement is also correct.
