If a recursive sequence converges, must its inverse be divergent? 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?

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.
