How to show sequence converges What I want to show is that the following sequence converges to 1.
$a_n = \frac{\sum_{i=0}^n i!}{n!} $
My initial strategy was to use the monotone sequence theorem. It's obvious that each term is positive, so it is bounded below by $0$. Plugging in this sequence into Desmos shows it appears to be decreasing (if $m>n$, then $a_m \le a_n$). Where I struggle is proving that it is decreasing. I've ran into roadblocks trying to show either $a_{n+1}-a_n \le0$ or $\frac{a_{n+1}}{a_n}\le 1$.
If I try to do the difference directly, I get stuck at showing $1-(1-\frac{1}{n+1})\frac{1}{n!}\sum_{i=0}^n i!\le 0$.
If I try and do the quotient, I get stuck at showing $\frac{n!}{\sum_{i=0}^ni!} +\frac{1}{n+1}<1$.
Is there something that I'm missing in the above attempts, or is there an easier way to show this converges using a different method?

Quotients are usually the way to go with factorials, but in this case the differences work, too.
$$a_n - (n+1)a_{n+1} = \frac{1}{n!}\left(\sum_{i=1}^{n}i!-\sum_{i=1}^{n+1}i!\right) = -(n+1)$$
$$\implies a_n-a_{n+1} = na_{n+1}-n-1$$
And
$$a_{n+1} = \frac{\sum_{i=1}^{n+1}i!}{(n+1)!} = 1 + \frac{\sum_{i=1}^{n}i!}{(n+1)!} > 1 + \frac{n\cdot n!}{(n+1)!} = 2 - \frac{1}{n+1} > 1 + \frac{1}{n}$$
which means
$$a_{n+1} > 1 + \frac{1}{n} \implies a_n-a_{n+1} > 0$$
