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?
 A: We can assume that $n\geq 2$. Note that
$$
\sum\limits_{i = 0}^n {i!}  = n! + (n - 1)! + \sum\limits_{i = 0}^{n - 2} {i!} 
$$
and
$$
\sum\limits_{i = 0}^{n - 2} {i!}  \le (n - 1)(n - 2)! = (n - 1)!.
$$
Thus,
$$
1 = \frac{{n!}}{{n!}} \le \frac{{\sum\nolimits_{i = 0}^n {i!} }}{{n!}} \le 1 + \frac{2}{n}.
$$
The limit now follows from the squeeze theorem.
A: 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
$$a_{n+1} > 1 + \frac{1}{n} \implies a_n-a_{n+1} > 0$$
A: It's a good start to show that this fraction is decreasing. As we have multiplication $\frac{a_{n+1}}{a_n}$ is the better candidate to look at:
$$
\frac{a_{n+1}}{a_n} = \frac{\sum_{i=0}^{n+1} i!}{(n+1)!} \cdot \frac{n!}{\sum_{i=0}^n i!} = \frac{1}{n+1} \cdot \frac{\sum_{i=0}^{n+1} i!}{\sum_{i=0}^n i!} = \frac{1}{n+1} \cdot \left(\frac{\sum_{i=0}^{n} i!}{\sum_{i=0}^n i!} + \frac{(n+1)!}{\sum_{i=0}^n i!}\right)
$$
Now we can make the last denominator smaller, by removing all summands but $n!$. We get:
$$
\frac{a_{n+1}}{a_n} < \frac{1}{n+1} \cdot \left(1 + \frac{(n+1)!}{n!}\right) = \frac{1}{n+1} \cdot \left(1 + (n+1)\right) = 1 + \frac{1}{n+1}
$$
I'm not sure if you can lower this bound to $1$, but I don't think this is needed.
If you now find a similar lower bound you can use the squeezing lemma to see that for $n\to\infty$ $\frac{a_{n+1}}{a_n} \to 1$.
