If $s(1)=1$ and $s(n)=s(n-1)+\text{lcm}[n,s(n-1)]-(-1)^n$ then $\lim\limits_{n\to \infty}\frac{s(n)}{(n+1)!}=\frac{1}{e}$ 
Define the sequence $(s(n))$ recursively by $s(1)=1$ and, for every $n\ge2$,
  $$s(n)=s(n-1)+\text{lcm}[n,s(n-1)]-(-1)^n.$$ Prove that
  $$\lim_{n\to \infty}\frac{s(n)}{(n+1)!}=\frac{1}{e}$$

I got this idea from a paper written by Eric Roland on recurrence relation on generating primes, the last part of the paper is where the actual idea comes from (Benoit Cloitre).
We noticed that this sequence converges to $e$ but we do not know how to prove it.
 A: First of all let's prove that $(n,s(n-1)) = 1$, by induction. Obviously the first few base case hold by simple calculation. Now let's assume that $(k,s(k-1)) = 1$, for some $k \in \mathbb{N}$. Now assume that $p$ is a prime divisor of $k+1$, then we have:
$$s(k) \equiv s(k-1) + [k,s(k-1)] - (-1)^k \equiv (k+1)s(k-1) - (-1)^k \equiv (-1)^k \pmod p$$
So from this follows that $p\nmid s(k) \implies (k+1, s(k)) = 1$. Hence the proof.
Now with a little bit backtracking we obtain for $n$ odd that:
$$\frac{s(n)}{(n+1)!} = \frac{s(n-2)}{(n-1)!} - \frac{n}{(n+1)!} = \frac{s(n-2)}{(n-1)!} - \frac{1}{n!} + \frac{1}{(n+1)!}$$
$$ = \frac{s(n-4)}{(n-3)!} - \frac{1}{(n-2)!} + \frac{1}{(n-1)!} - \frac{1}{n!} + \frac{1}{(n+1)!} = \frac{s(1)}{2!} + \sum_{i=3}^{n+1} \frac{(-1)^i}{i!} = \sum_{i=0}^{n+1} \frac{(-1)^i}{i!}$$
Similarly one can obtain the same formula for $n$ even (again we'll have minus in front of odd terms and plus in front of even terms). So now:
$$\lim_{n\to \infty} \frac{s(n)}{(n+1)!} = \lim_{n\to \infty} \sum_{i=0}^{n+1} \frac{(-1)^i}{i!} = \frac 1e$$
As the LHS is actually the Taylor series expansion of $f(x)=e^x$, evaluated at $x=-1$.
