TITLE: Suppose $(a_n)$ is a sequence such that $a_n=\frac{1!+2!+\cdots+n!}{n!}$. Show that $\lim{a_n}=1$
QUESTION [11 upvotes]: Suppose $(a_n)$ is a sequence such that $$a_n=\frac{1!+2!+\cdots+n!}{n!} \, .$$ Show that $\lim_{n \rightarrow \infty}{a_n}=1$. My attempt is to formulate an inequality and then use the Squeeze Theorem. Since we know that $$\frac{1}{n!}+\frac{2!}{n!}+\cdots+1>1$$ as $\frac{i!}{n!}>0$ for all $1 \leq i \leq n-1$, we have $a_n \geq1$. Then I'm stuck at formulating another side. Can anyone help?

REPLY [6 votes]: 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.