does this nested radicle converge? $1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}$ Let $a_0=1$ and $a_n=1+\sqrt{a_{n-1}}+\sqrt{1+\sqrt{a_{n-1}}}$
I want to know if the limit of $a_n$ as n goes to infinity.
$$1+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}}+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}}}$$
When I tried to  find a possible answer I tried to solve $1+\sqrt{x}+\sqrt{1+\sqrt{x}}=x$. I tried for an hour so I gave up and asked WolframAlpha and it got $x≈5.04891733952231$
I'm not sure how to find out if $a_n$ converges. I think it does

Starting from @Will Jagy first hint $$w^3 - 2 w^2 - w + 1 = 0\qquad \text{with} \qquad x=w^2$$ the cubic shows three real roots $(\Delta=49)$. Using the trigonometric method, they are
$$w_k=\frac{2}{3} \left(1+\sqrt{7} \cos \left(\frac{1}{3} \left(2 \pi  k-\sec ^{-1}\left(2
   \sqrt{7}\right)\right)\right)\right)\qquad \text{with} \qquad k=0,1,2$$ and the solution of concern is $w_0$ (not prooved).
This gives
$$x=\frac{4}{9} \left(1+\sqrt{7} \cos \left(\frac{1}{3} \sec ^{-1}\left(2
   \sqrt{7}\right)\right)\right)^2$$ which is another expression of the result.
