# $a_m(j+1)=\left(a_m(j)\right)^2+2a_m(j) \space\space(j\ge 0)$: Evaluate $\lim_{n \to \infty}a_n(n)$

Question_

$$d$$ is a real number. In each integer $$m\ge0$$, we define $$\{a_m(j)\}$$ $$(j=0, 1, 2, \cdots)$$ as follows. $$a_m(0)=\frac{d}{2^m}, a_m(j+1)=\left(a_m(j)\right)^2+2a_m(j) \space\space(j\ge 0)$$ Then, evaluate $$\lim_{n \to \infty}a_n(n)$$.

It is known that the closed-form of the given sequence is $$a_m(j)+1 = (a_m(0)+1)^{2^{j-1}} \space\space\cdots(1)$$ It is not that hard to find $$\lim_{n \to \infty}a_n(n)$$ from here.

My question is that how the closed-form of the given sequence can be calculated like that?

For sure, it can be proven by mathematical induction. However, what I do want to know is how the closed-form is derived, not how the $$eq.(1)$$ is fit to the given recursion formula. Thanks for answering!

• I think the way is to apply recurrence relation to guess the formula and then prove the formula by induction. You can read something here: en.wikipedia.org/wiki/Recurrence_relation – asv 2 days ago

From the recurrence relation, you can get $$a_m(j+1) + 1 = (a_m(j)+1)^2$$. Then, you can get the closed-form by using this relation recursively.