Suppose $(a_n)$ is a sequence such that $a_n=\frac{1!+2!+\cdots+n!}{n!}$. Show that $\lim{a_n}=1$

Solution 1:

Each of $1!,2!,...,(n-2)!$ is at most $(n-2)!$. Since there are $n-2$ such terms you have $$\frac{1! + 2! + \cdots + n!}{n!} \leq \frac{(n-2)(n-2)! + (n-1)! + n!}{n!}$$ $$= \frac{n-2}{n(n-1)} + \frac{1}{n} + 1$$ The limit of this as $n \rightarrow \infty$ is $1$, which is what you need.

Solution 2:

We will prove that: $$1!+2!+\cdots+n!\le 2\cdot n!$$ If $n=1$, it is trivial. And if $1!+2!+\cdots+n!\le 2\cdot n!$ holds, then $$1!+2!+\cdots+n!+(n+1)! \le 2\cdot n! +(n+1)! =(n+3)n! \le 2(n+1) n!$$ for all $n\ge 1$.

So we get $$\frac{1!+2+\cdots+(n-1)!+n!}{n!}\le \frac{2(n-1)!+n!}{n!}=1+\frac{2}{n}.$$

Solution 3:

As Jonas Meyer suggested, the Stolz–Cesàro theorem might be another option. $$\lim_{n\to\infty}\frac{1!+2!+\cdots+n!}{n!}=\lim_{n\to\infty}\frac{(n+1)!}{(n+1)!-n!}=\lim_{n\to\infty}\frac{n+1}{n}=1.$$

Done.