<s>[INST] Recursive sequence of functions Recursive sequence of functions: $f_{n+1}= \sqrt{x+f_n}$, $f_1(x)= \sqrt{x}$. 
this sequence is monotonic, but what is bounding it? thanks 
 [/INST] So you have defined $$f_n(x) = \underbrace {\sqrt{x + \sqrt {x + \dots \sqrt x}}}_{\text {n square roots}}$$ and you say that it is "monotonic", suggesting some comparison like
$$f_{n+1} \ge f_n$$
which, correct me if I'm wrong, I'm going to assume means $\forall x\, f_{n+1}(x) > f_n(x)$.
A bound (..but not a least upper bound...) then would imply that you are looking for a function $g(x)$ such that
$$\forall n,\,x,\, g(x) \ge f_n(x)$$
This is a bit more than I would expect in a regular infinitessimal calculus class.  The upper bound obviously has to be larger than $g(x) = \sqrt x$, but it doesn't seem like it has to be much more.  Let's try $g(x) = \sqrt {2x}$.  To use induction though, we have to prove a stronger result:
$$h_n(x) = \sqrt{\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^{n-1}}\right)\cdot x}$$
$$h_n(x) \ge f_n(x)$$
Base case:
$$\sqrt{x} \ge \sqrt{x}$$
Recursive assumptions:
$$h_n(x) \ge f_n(x)$$
Recursive cave:
$$h_{n+1}(x) \ge f_{n+1}(x)$$
$$\sqrt{\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^n}\right)\cdot x} \ge \sqrt{x + f_n(x)}$$
$$\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^n}\right)\cdot x \ge x + f_n(x)$$
$$\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^{n - 1}}\right)\cdot x + \frac x {2^n} \ge x + f_n(x)$$
$$h_n(x)^2 + \frac x {2^n} \ge f_n(x) + x $$
$$\text{Is implied by}$$
$$f_n(x)^2 + \frac x {2^n} \ge f_n(x) + x $$
$$f_n(x)^2 - f_n(x) \ge x - \frac x {2^n} $$
This last statement....sorry...we have to prove with induction over n.
Base case:
$$x - \sqrt{x} > x - \frac x 1$$
For $x\ge 1$ this holds.
Recursive assumption:
$$f_n(x)^2 - f_n(x) \ge x - \frac x {2^n} $$
Recursive case:
$$f_{n+1}(x)^2 - f_{n+1}(x) \ge x - \frac x {2^{n+1}} $$
$$x+f_n(x) - \sqrt{x+f_n(x)} \ge x - \frac x {2^{n+1}} $$
$$f_n(x) - \sqrt{x+f_n(x)} \ge \frac x {2^{n+1}} $$
$$f_n(x) + \frac x {2^{n+1}} \ge \sqrt{x+f_n(x)}  $$
$$f_n(x)^2 + 2f_n(x)\frac x {2^{n+1}} + \frac {x^2} {2^{n+2}} \ge x+f_n(x)  $$
$$\text{Is implied by: (using the recursive assumption)}$$
$$x - \frac x {2^n} + 2f_n(x)\frac x {2^{n+1}} + \frac {x^2} {2^{n+2}} \ge x$$
$$x^2 - 4 x + 4f_n(x)\ge 0$$
Which holds for $x \ge 1$.
Since $\forall n\, g(x) \ge h_n(x)$, $g$ is an upper bound for all $f_n$, for $x \ge 1$. 
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