Evaluate $\lim\limits_{n \to \infty}\frac{1!+2!+\cdots+n!}{n!}$.

Question

Problem

Evaluate $$\lim\limits_{n \to \infty}\frac{1!+2!+\cdots+n!}{n!}.$$

Solution 1

Applying Stolz theorem, $$\lim_{n \to \infty}\frac{1!+2!+\cdots+n!}{n!}=\lim_{n \to \infty}\frac{(n+1)!}{(n+1)!-n!}=\lim_{n \to \infty}\frac{n!(n+1)}{n!\cdot n}=\lim_{n \to \infty}\frac{n+1}{n}=1.$$

Solution 2

\begin{align*} 1=\frac{n!}{n!}\leq \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\to 1(n \to \infty). \end{align*} Applying the squeeze theorem, $$\lim\limits_{n \to \infty}\frac{1!+2!+\cdots+n!}{n!}=1.$$ Any more solutions? Thanks.

Answer 1

This is not much different from the second one, but perhaps the idea is slightly more methodical. Call $a_n=\frac{1}{n!}\sum_{j=0}^n j!$. Then $a_{n}=\frac1{n}a_{n-1}+1$. Since $a_0=1$, the whole sequence is positive and bounded (specifically, $1\le a_n\le 2$). Therefore, $$\limsup_{n\to\infty}a_n\le \limsup_{n\to\infty}\frac2n+1=1$$.