Suppose I have a recursive sequence $\displaystyle a_{n+1} = \frac{a_{n}}{2}$. Clearly, the sequence converges towards zero. Now, suppose I define an "inverse" sequence $\displaystyle b_{n+1} = 2b_{n}$. While the sequence clearly diverges, could I know that based on the knowledge that $a_{n+1}$ converges? Is there any sequence that converges, whose "inverse" also converges?
.. and could this "inverse" sequence be defined in a more intuitive way?
On
Consider the sequence $a_{k+1} = \sinh(a_k)$. If started at $a_0=\exp(i \pi/4)$ then this converges to $\lim_{n \to \infty} a_n=0$.
Now look at the inverse $ a_{k-1} = \operatorname{asinh}(a_k)$. you can step backwards to arrive at $a_0$ again and go to $\lim_{n \to -\infty} a_n=0$
When I came across it this surprised me much. See a picture here: 
Say the sequence $(a_n)$ defined by $a_{n+1} = f(a_n)$ for bijective $f$ converges to $a$.
If $f$ is continuous at $a$, then $$0 = \lim\limits_{n \to \infty} a_ {n+1}-a_n = \lim\limits_{n \to \infty} f(a_n)-a_n = f(a)-a$$
So $f(a)=a$. But then it also follows that $f^{-1}(a)=a$. So if the inverse sequence is started with seed value $b_0 = a$, the inverse sequence will converge also.
I'm not $100$% sure what can be said for such convergent recurrences where $f$ is not continuous at the limit.