Limit of recursive sequence $x_{n} =(x_{n-1})^b(ab+a)-1$

Question

I have the following infinite series, of which the following are the $1^{st}, 2^{nd}$, $3^{rd}$ and $4^{th}$ term

$$2$$

$$\big(2\big)^{b}(ab+a)-1$$

$${\bigg(\big(2\big)^{b}(ab+a)-1)\bigg)^b(ab+a)-1}$$

$$\Bigg({\bigg(\big(2\big)^{b}(ab+a)-1)\bigg)^b(ab+a)-1}\Bigg)^b(ab+a)-1$$

$$etc.$$

The pattern seems quite straightforward, the $n^{th}$ (i.e. $x_{n}$) simply is

$$x_{n} =(x_{n-1})^b(ab+a)-1$$

I want to come up with a formula that describes the $n^{th}$ term (It's probably hard to define it for $n\geq0$, i dont mind if it only holds for $n\geq1$.)

Second, I want to find the limit as $n->\infty$. I think this requires the series to be convergent- some tips would be welcome.

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values of a and b: $a>0,b>1$

Answer 1

If this sequence has a limit $L$, then $L = L^b(ab+a)-1$.

There isn't much hope of an explicit formula for $L$.

Indeed, for $b=5$ most probably there won't be a formula for $L$ using elementary operations.