How can I prove this equality and general derivative? $$ f_{n}(x) = \underbrace{\sqrt{x+\sqrt{x+ ...+ \sqrt{x}}}}_{n}  \quad \quad \lim_{n\to\infty} \sum^{n}_{r=1}\frac{1}{2^{r}}\prod^{r-1}_{i=0}\frac{1}{f_{n-i}(x)} = \frac{1}{2f_{\infty}(x)-1} $$
I produced the left hand side by examining the pattern of the derivatives of $\, f_{n}(x)$ and the right hand side by using the property of $\, f_{\infty}(x)$ that: $\, f_{\infty}(x)=\sqrt{x + f_{\infty}(x)}$ and differentiating. I have no idea how to prove such a thing, nor do I know how to prove the general formula of the derivate (EDIT: I have now proved the general formula of the derivative but am still stumped by the limit problem!):
$$_{m}f_{n}(x) = \underbrace{\sqrt[m]{x+\sqrt[m]{x+ ... + \sqrt[m]{x}}}}_{n}  \quad \quad
_{m}f_{n}'(x) = \sum^{n}_{r=1}\frac{1}{m^{r}} \prod^{r-1}_{i=0} (f_{n-i}(x))^{1-m}$$
Obviously given my second formula,  I could generalise the first equality but, I thought the case $m=2$ came out quite nicely. 
One last thing, is there any nicer notation I can use for the function rather than writing it out with the underbrace and whatnot?
Edit: Just realised I didn't mention it at all in the original post, the right-hand side is the infinite sum of a geometric progression:
$$\sum^{\infty}_{r=1}\bigg(\frac{1}{2f_{\infty}(x)}\bigg)^{r} = \frac{1}{2f_{\infty}(x)-1} $$
So the question can be rephrased as proving:
$$\lim_{n\to\infty} \sum^{n}_{r=1}\frac{1}{2^{r}}\prod^{r-1}_{i=0}\frac{1}{f_{n-i}(x)} = \sum^{\infty}_{r=1}\bigg(\frac{1}{2f_{\infty}(x)}\bigg)^{r}$$
Which seems intuitively true but, I do not know how to prove it rigorously. 
 A: It took me a while to see how you managed to find that formula, but indeed when you define
$$
_mf_n(x) = (x + {_mf_{n-1}}(x))^{1/m}
$$
with $f_{n} = 0$ for $n \leq 0$, then it follows that
$$
_mf'_n(x) = \frac1m(x + {_mf'_{n-1}}(x))^{1/m - 1} (1 + {_mf'_{n-1}}(x))
= \frac1m(_mf_n(x))^{1 - m} (1 + {_mf'_{n-1}}(x))
$$
and through induction you can then show that
$$
_{m}f_{n}'(x) = \sum^{n}_{r=1}\frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m}.
$$
actually if we let $(_mf_{n}(x))^{1-m} = 0$ for $n \leq 0$ then we can make our lives slightly easier by writing this as
$$
_{m}f_{n}'(x) = \sum^{\infty}_{r=1}\frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m}.
$$
If we now let $_mf_{\infty}(x) = \lim_{n \to \infty} {_mf_n}(x)$ then we can show that for any $r$
$$
\lim_{n\to\infty} \frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m} = \left(\frac{(_mf_{\infty}(x))^{1-m}}{m} \right)^{r}
$$
as long as $_mf_{\infty}(x)$ exists.
We can also show that $_mf_{n}(x)$ is monotonically increasing in $n$. Note that for all $x > 0$ we have ${_mf_{1}}(x) = x^{1/m} > 0 = {_mf_{0}}(x)$ and if
$$
{_mf_{n}}(x) \geq {_mf_{n-1}}(x)
$$
then
$$
{_mf_{n+1}}(x) = (x + {_mf_{n}}(x))^{1/m} \geq (x + {_mf_{n-1}}(x))^{1/m} = {_mf_{n}}(x)
$$
so ${_mf_{n}}(x)$ is always monotonically increasing for $x>0$. This allows us to use the monotone convergence theorem to conclude that for all $x>0$
$$
\begin{align}
\lim_{n\to\infty} \sum^{\infty}_{r=1}\frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m} 
&= \sum^{\infty}_{r=1} \lim_{n\to\infty} \frac{1}{m^{r}} \prod^{r-1}_{i=0} (_mf_{n-i}(x))^{1-m}\\
&= \sum^{\infty}_{r=1} \left(\frac{(_mf_{\infty}(x))^{1-m}}{m} \right)^{r}\\
&= \frac{1}{1 - \frac{(_mf_{\infty}(x))^{1-m}}{m}} - 1\\
&= \frac{(_mf_{\infty}(x))^{1-m}}{(_mf_{\infty}(x))^{1-m} - m}
\end{align}
$$
setting $m=2$ results in the special case you found.
You could also try to prove this by showing
$$
\lim_{n\to\infty} \lim_{h\to0} \frac{_{m}f_{n}(x+h) - {_{m}f_{n}(x)}}{h} =
\lim_{h\to0} \lim_{n\to\infty} \frac{_{m}f_{n}(x+h) - {_{m}f_{n}(x)}}{h} 
$$
but showing that you're allowed to exchange the two limits is a bit involved, especially with no idea what the function looks like. So using the expression you found for $_{m}f_{n}'(x)$ might well be the easiest approach.
