After exploring Ramanujan's famed $$3=\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots}}} $$ and $$4=\sqrt{6+2\sqrt{7+3\sqrt{8+\cdots}}},$$ both of which can be expressed more generally by $$x+n+a=\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{a(x+2n)+\cdots}}},$$ I came across a nested radical of the form \begin{equation}P=\sqrt{a+\sqrt{ab+\sqrt{ab^2+\sqrt{ab^3+\cdots}}}} \end{equation} and am struggling to derive a formula for determining to what value $P$ converges. I know from a simple reformulation of $$\sqrt{a+b\sqrt{a+b\sqrt{a+\cdots}}}$$ that $$\sqrt{a+\sqrt{ab^2+\sqrt{ab^6+\sqrt{ab^{14}+\cdots}}}} $$ converges to $$\frac{b+\sqrt{b^2+4a}}{2}. $$ This result, however, fails to be helpful. Any advice on determining the convergence of $P$ would be appreciated.
UPDATE:
Let \begin{equation} P=\sqrt{a+\sqrt{ab+\sqrt{ab^2+\sqrt{ab^3+\cdots}}}}.\end{equation} Notice $$P(x)=\sqrt{ab^x+P(x+1)} $$ denotes the recurrence relation for $P$. My new attempt was to consider Ramanujan's recurrence relation: $$\psi(x)=\alpha+\beta+x=\sqrt{\alpha x +(\alpha+\beta)^2+x\psi(x+1)}, $$ let $\beta=1$, then set \begin{equation}ab^x=\alpha x +(\alpha+1)^2.\end{equation} Because $\alpha\geq0$, the above produces $$\alpha=\frac{-(x+2)+\sqrt{(x+2)^2-4(1-ab^x)}}{2}. $$ Through Ramanujan's formula, setting $x=1$ then implies that $$\psi(1)=\frac{1+\sqrt{5+4ab}}{2}=\sqrt{ab+\psi(2)}.$$ I realize, as suggest in Infinitely nested radical problem?, which is nearly identical to my new approach, that $P$ and $\psi$ are not in complete equitable forms. That post also suggests that this may not be the correct approach to solving this problem. Could someone in different words than the cited post please explain why this method is not sufficient and perhaps propose a new method or resource for computing radicals of this form?
