Convergence of a finite continued fraction

Question

Given a sequence of natural numbers $(a_n)_{n\geq 1}$, we know that the finite continued fraction $$[a_1,a_2,\ldots,a_n]:=\cfrac{1}{a_1+\cfrac{1}{\ddots+\cfrac{\ddots}{a_{n-1}+\cfrac{1}{a_n}}}}$$ is convergent when $n\to\infty$. My question is: what about the convergence of the reverse finite continued fraction $[a_n, a_{n-1},\ldots, a_1]$?

Answer 1

Not necessarily. An easy counterexample: $a_n = 1$ if $n$ is even, $2$ if $n$ is odd.

Answer 2

Intuitively, it should be clear that the reverse variant does not converge. The point is that for the normal case, "the influence of a new number" vanishes in the limit, while in the reverse case "the last number" has a massive influence.
Just take the sequence $K_n = [1,2,1,2,1,2,1,2......]$ such that $K_n$ has $n$ elements (cardinality). Consider these two for the normal case: \begin{equation} [1,2,1,2,1....] \to \frac{1}{1+\frac{1}{2+\frac{1}{1 + ...}}}\end{equation} \begin{equation}[2,1,2,1,2....] \to \frac{1}{2+\frac{1}{1+\frac{1}{2 + ...}}}\end{equation} These are not equal. Considering this, we see that in the reverse case, $K_n$ and $K_{n+1}$ would always be different, and won't converge. To make it more extreme you could use a larger number, instead of 2.