# 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?

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?

• I've added parentheses around the $n+1$ in the denominator as this looks to be your intent based on the Desmos link. – Michael Lugo Jul 18 '18 at 19:26
• @MichaelLugo Yes, thank you, I tend to simplify to the point of my own detriment – user189728 Jul 18 '18 at 19:28

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 $$\lim a_n=-1$$
• Dear layman, did you try to find a relation between $b_{n+1}$ and $b_n$? – Did Jul 18 '18 at 19:34