A sequence with infinitely many radicals: $a_{n}=\sqrt{1+\sqrt{a+\sqrt{a^2+\cdots+\sqrt{a^n}}}}$

Here is a full answer to part (a) and a partial answer to part (b). Call $(a_n(a))_{n\geqslant1}$ the sequence when the value of the parameter is $a$.

One has $a_0(1)=1$ and $a_{n+1}(1)=u(a_n(1))$ for every $n\geqslant0$ with $u(x)=\sqrt{1+x}$. Hence the usual technique shows that the sequence $(a_n(1))_{n\geqslant0}$ is increasing to $a_\infty(1)=\alpha$ where $\alpha$ solves the equation $\alpha=u(\alpha)$, that is, $\color{red}{\alpha=\frac12(1+\sqrt5)}$.

When $a\lt1$, $a_n(a)\leqslant a_n(1)$ and $(a_n(a))_{n\geqslant0}$ is increasing hence $(a_n(a))_{n\geqslant0}$ converges to a finite limit $a_\infty(a)$ with $\color{red}{\sqrt{1+\sqrt{a}}\lt a_\infty(a)\leqslant \alpha}$.

When $a\gt1$, $\sqrt{1+\sqrt{aa_{n-1}(1)}}\leqslant a_n(a)\leqslant\sqrt{1+\sqrt{a}a_{n-1}(1)}$ and $(a_n(a))_{n\geqslant0}$ is increasing hence it converges to a finite limit $a_\infty(a)$ with $\color{red}{\sqrt{1+\sqrt{\alpha a}}\lt a_\infty(a)\leqslant\sqrt{1+\alpha\sqrt{a}}}$.

To show the upper bound on $a_n(a)$, one carries over every power of $a$ to the left until it reaches the position of $\sqrt{a}$. Crossing a square root sign halves the exponent and $a\gt1$ hence the power of $a$ which just crossed a square root sign is smaller than the preceding one. For example, the first step of the proof uses $a^{n/2}\leqslant a^{n-1}$ to deduce $$ \sqrt{a^{n-1}+\sqrt{a^n}}=\sqrt{a^{n-1}+a^{n/2}\sqrt1}\leqslant\sqrt{a^{n-1}(1+\sqrt1)}=a^{(n-1)/2}\sqrt{1+\sqrt1}, $$ the second step uses $a^{(n-1)/2}\leqslant a^{n-2}$, and so on, until $a^{3/2}\leqslant a^2$ and $a^{2/2}\leqslant a$. A similar reasoning yields the lower bound.

Finally, the map $a\mapsto a_\infty(a)$ is nondecreasing from $\color{red}{a_\infty(0)=1}$ to $\color{red}{a_\infty(+\infty)=+\infty}$.

The Herschfeld's Convergence Theorem may help (the more general subject is 'Nested Radical').

Herschfeld's paper "On Infinite Radicals" is available here.