Does the limit of the recursive sequence $a_{n+1}=\frac{n+1}{\frac{1}{a_{n}}-(n+1)}$ converge to the same value regardless of whatever $a_0$ is?

Question

I found an interesting recursive sequence $a_{n+1}=\frac{n+1}{\frac{1}{a_{n}}-(n+1)}$. In playing with $a_0$ on Desmos, it seems that no matter what $a_0$ is, the sequence converges to some fixed value. Is this the case? Is this sequence the same regardless of its initial value? If so, what is the limit of this sequence?

Answer 1

The sequence $$b_n=\frac{n!}{a_n}$$ solves the recursion $$b_{n+1}=b_n-(n+1)!$$ hence $$b_n=b_0-\sum_{k=1}^nk!$$ which yields $$\frac1{a_n}=\frac1{n!a_0}-\sum_{k=1}^n\frac{k!}{n!}$$ Now, for every $n\geqslant2$, $$1\leqslant\sum_{k=1}^n\frac{k!}{n!}\leqslant1+\frac1n+\frac{n-2}{n(n-1)}$$ Thus, for every initial condition $a_0$ not in the set of the sums $\sum\limits_{k=1}^nk!$ for $n\geqslant0$, one has $$\lim a_n=-1$$