How would one evaluate the infinitely nested radical $$\sqrt{x+\sqrt{\sqrt{x+\sqrt{\sqrt{\sqrt{x+...}}}}}}$$ due to the fact that each term cannot be evaluated individually to create a product or sum, and using repeated difference of two squares to find a pattern also doesn't seem to work, I am not sure where to even start. Any solutions or help would be appreciated
$$\text{it can be written as follows to make the pattern more obvious}$$ $$f(x) = \left(x+\left(x+\left(x+...\right)^{\frac{1}{8}}\right)^{\frac{1}{4}}\right)^{\frac{1}{2}}$$ and defined by the limit of the recursive sequence of functions (credit to @Andrea Marino) $$ \left\{\begin{array}{rcl} f_1(x,y) & = & y^{1/2}\\ f_n(x,y) & = & f_{n - 1}(x,x + y^{1/2^n}) \textrm{ if } n \geqslant 1 \end{array}\right. $$
update: This questions is bountied for anyone who can find a closed form, or at least an accurate approximation
Your problem seems very interesting, I spent a bit of time on it and here is what I got. First of all, as said @Andrea Marino, this infinite root can be defined as the limit of the recursive sequence of functions, $$ \left\{\begin{array}{rcl} f_1(x,y) & = & y^{1/2}\\ f_n(x,y) & = & f_{n - 1}(x,x + y^{1/2^n}) \textrm{ if } n \geqslant 1 \end{array}\right. $$ Here, $y$ represents the starting point of your sequence of roots. Andrea suggested that we start at $y = x$ but we will see actually that $f_n$ converges to a limit which doesn't depend on $y$.
Indeed, some numerical observations suggest that, \begin{array}{l} *\ f_n(x,y) \rightarrow f(x) \in \mathbb{R}_+ \textrm{ simply},\\ *\ \textrm{ The convergence is uniform on each } [\varepsilon,+\infty[,\\ *\ \lim_{x \rightarrow 0^+} f(x) = 1,\\ *\ f \textrm{ is smooth on } ]0,+\infty[,\\ *\ f \textrm{ increases}. \end{array} Notice that each $f_n$ is clearly increasing with respect to both $x$ and $y$ so the last point is trivial if we assume that $f$ exists.
To get easily the smoothness (and even the analyticity !), we can extend the domain of the $f_n$ on the whole $\mathbb{H}^2$ where $\mathbb{H} = \{z \in \mathbb{C}|\Re(z) > 0\} = \{z \in \mathbb{C}|-\pi/2 < \arg(z) < \pi/2\}$. Indeed, we can extend $m$-th roots on this set by $(re^{i\theta})^{1/m} = r^{1/m}e^{i\theta/m}$ where $\theta$ is chosen between $-\pi/2$ and $\pi/2$. This defined holomorphic functions from $\mathbb{H}$ to itself. Moreover, $\mathbb{H}$ is stable by sum, thus the $f_n$ are well defined since $(x,y) \in \mathbb{H}^2 \Rightarrow x + y^{1/2^n} \in \mathbb{H}$. We can even choose $x \in \overline{\mathbb{H}} = \{z \in \mathbb{C}|\Re(z) \geqslant 0\}$.
The purpose of doing this is to define $f$ on the whole $\overline{\mathbb{H}}$ instead of $\mathbb{R}_+^*$ and to prove in the same time that it is holomorphic using theorems like Morera's theorem that is useful to show that a limit of holomorphic functions is holomorphic, thus we should obtain the highest possible level of smoothness for $f$. Finally, we can study the limit at $x = 0$.
Lipschitzianity of the $f_n$
First of all, let $\mathbb{H}_\varepsilon = \{z \in \mathbb{H}|\Re(z) \geqslant \varepsilon\}$. Let us prove that $f_n(x,\cdot)$ is $k_n(\varepsilon)$-Lipschitzian with respect to $y$ in $\mathbb{H}_\varepsilon$ where $k_n(\varepsilon)$ will be determined later. Recall that on $\mathbb{H}_\varepsilon$, $$ \left|\frac{d}{dy}y^{1/m}\right| = \left|\frac{1}{m}y^{\frac{1}{m} - 1}\right| \leqslant m^{-1}\varepsilon^{-1 + 1/m}, $$ thus $y \mapsto y^{1/m}$ is $m^{-1}\varepsilon^{-1 + 1/m}$-Lipschitzian so in particular, $f_1(x,\cdot)$ is $\frac{1}{2\sqrt{\varepsilon}}$-Lipschitzian i.e. $k_1(\varepsilon) = \frac{1}{2\sqrt{\varepsilon}}$ fits.
Now, if $y = re^{i\theta} \in \mathbb{H}_\varepsilon$ with $-\pi/2 < \theta < \pi/2$, $y^{1/m} = r^{1/m}e^{i\theta/m}$. We deduce that $\Re(y) = r\cos(\theta) \geqslant \varepsilon$ so $r \geqslant \varepsilon$ and $$ \Re(y^{1/m}) = r^{1/m}\cos\left(\frac{\theta}{m}\right) \geqslant r^{1/m - 1}r\cos(\theta) \geqslant \varepsilon^{1/m - 1}\varepsilon = \varepsilon^{1/m}. $$ Therefore, $x + y^{1/m} \in \mathbb{H}_{\varepsilon^{1/m}}$ hence by induction, \begin{align*} |f_n(x,y) - f_n(x,z)| & = |f_{n - 1}(x,x + y^{1/2^n}) - f_{n - 1}(x,x + z^{1/2^n})|\\ & \leqslant k_{n - 1}(\varepsilon^{1/2^n})|y^{1/2^n} - z^{1/2^n}|\\ & \leqslant k_{n - 1}(\varepsilon^{1/2^n})2^{-n}\varepsilon^{-1 + 1/2^n}|y - z|. \end{align*} It show that indeed $f_n(x,\cdot)$ is $k_n(\varepsilon)$-Lipschiztian with $k_n(\varepsilon) = k_{n - 1}(\varepsilon^{1/2^n})2^{-n}\varepsilon^{-1 + 1/2^n}$. Let us show by induction that we have the bound $k_n(\varepsilon) \leqslant 2^{-n(n + 1)/2}\varepsilon^{-1}$ when $\varepsilon \leqslant 1$. Indeed, $k_1(\varepsilon) = 2^{-1}\varepsilon^{-1/2} \leqslant 2^{-1}\varepsilon^{-1}$ and for all $n \geqslant 2$, \begin{align*} k_n(\varepsilon) & = k_{n - 1}(\varepsilon^{1/2^n})2^{-n}\varepsilon^{-1 + 1/2^n}\\ & \leqslant 2^{-(n - 1)n/2}(\varepsilon^{1/2^n})^{-1}2^{-n}\varepsilon^{-1 + 1/2^n}\\ & \leqslant 2^{-n(n + 1)/2}\varepsilon^{-1}\\ \end{align*} It shows in particular that $f_n(x,\cdot)$ is $\frac{1}{\varepsilon2^{n(n + 1)/2}}$-Lipschitzian which is a good sign that the limit of the $f_n$, if it exists, won't depend on $y$.
Convergence of $f_n(\cdot,1)$
By Lipschitzianity, we have that for all $n$ and all $x$, $$ |f_n(x,1) - f_{n - 1}(x,1)| = |f_{n - 1}(x,x + 1) - f_{n - 1}(x,1)| \leqslant \frac{|x|}{2^{(n - 1)n/2}}. $$ We deduce that $\sum f_n(\cdot,1) - f_{n - 1}(\cdot,1)$ converges normally on all compact subset of $\overline{\mathbb{H}}$ because $\sum 2^{-(n - 1)n/2}$ converges. In particular, $f_n(x,1) \rightarrow f(x)$ when $n \rightarrow +\infty$ locally uniformly. Local uniform convergence preserves continuity and holomorphicness (it follows from Morera's theorem) thus $f$ is continuous on $\overline{\mathbb{H}}$ and holomorphic on $\mathbb{H}$. Now, if we take a $y \in \mathbb{H}_\varepsilon$ for a $0 < \varepsilon \leqslant 1$, we have, \begin{align*} |f_n(x,y) - f(x)| & \leqslant |f_n(x,y) - f_n(x,1)| + |f_n(x,1) - f(x)|\\ & \leqslant \frac{|y - 1|}{\varepsilon 2^{n(n + 1)/2}} + |f_n(x,1) - f(x)|\\ & \rightarrow 0, \end{align*} and the convergence is uniform if we take $x$ in a compact subset of $\overline{\mathbb{H}}$ and $y$ in a $\mathbb{H}_\varepsilon$ (in particular, we have uniform convergence on all compact subset of $\overline{\mathbb{H}} \times \mathbb{H}$). The limit doesn't depend on the choice of $y$.
Discussion
$*$ As we must have $\Re(y) > 0$, then for all $n$, $f_n(0,y) = y^{1/2^{n(n + 1)/2}} \rightarrow 1$ so $f(0) = 1$ and not $0$ as the expression with the imbricated square roots sugests it.
$*$ What happens if we replace $f_n(x,y) = f_{n - 1}(x,x + y^{1/2^n})$ by $f_n(x,y) = f_n(x,x + y^{a_n})$ for some sequence $a_n \in [0,1]$ ? In the proof of the convergence of $(f_n(\cdot,1))$, we needed the fact that $\sum_{n \geqslant 0} 2^{-n(n + 1)/2} < +\infty$ which probably becomes in this generalized case $\sum_{n \geqslant 0} \prod_{k = 1}^n a_k < +\infty$.
$*$ Is is possible to continue $f$ on a larger domain of $\mathbb{C}$ ? Like for power functions, I think we can't continue $f$ on the whole $\mathbb{C}$ but maybe on any domain that doesn't contain $0$ and in which any loop has a null index around $0$.