I'm trying to prove that this sequence $(x_n)$, where $x_1 =\sqrt a$ and $x_{n+1}=\sqrt{a +x_n}$ has a limit, then I would like to find the limit of
$L=\sqrt {a+\sqrt{a+\sqrt{a+\sqrt{a...}}}}$
It's easy to find this limit and prove that this sequence is monotone (induction over $\mathbb N$). What I found difficult is prove that this sequence is bounded.
I need help in this part.
Thanks a lot.
When $a\ge1$, we have $x_1=\sqrt{a}\le a<2a$ and by mathematical induction, $$x_{n+1}^2 = a + x_n < a+2a=3a < 4a^2\ \Rightarrow\ x_{n+1}<2a.$$ When $a<1$, the sequence is dominated by the analogous sequence with $a=1$ and hence it is bounded.
If the limit exists, it is obvious to see that $$ \lim_{n\rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty} a_n$$ So lets call the limit $x$ than $$x=\sqrt{a+x}$$ For the bound, i would use banach fix point theorem. The fix point theorem together with the fact, that the root is a strict contraction. So we know $$x_1\leq \sqrt{a} + \sqrt{\sqrt{a}}$$ For $a$ sufficiently large $a$ will be a upper bound and $0$ a lower bound. For $a$ not sufficiently large we don't need a proof, since we know it is lower bounded by $0$ and monotone increasing, so we can take the upper bound of the sufficiently large $a$ (monotonicity of the root).
