Lots of $x$ th roots of $x$

$\begin{aligned} \sqrt[x]{x\sqrt[x]{x\sqrt[x]{x\cdots}}} & = x^x & \small \color{#3D99F6} \text{Raise both sides to the power of }x \\ x \sqrt[x]{x\sqrt[x]{x\sqrt[x]{x\cdots}}} & = \left(x^x\right)^x \\ x\cdot x^x & = x^{x^2} \\ x^{x+1} & = x^{x^2} \end{aligned}$

We note that $x=1$ is a solution and equating the exponents or powers on both sides, we have:

$\begin{aligned} x+1 & = x^2 \\ x^2 - x - 1 & = 0 & \small \color{#3D99F6} \text{Since }x > 0 \\ \implies x & = \frac {1+\sqrt 5}2 = \color{#3D99F6} \varphi & \small \color{#3D99F6} \text{where }\varphi \text{ is the golden ratio.} \end{aligned}$

Therefore the sum of solutions is $1+\varphi \approx \boxed{2.6180}$ .

We can write the equation as the following: $x^{\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}...+\frac{1}{x^\infty}}=x^x$

Notice that 1 and 0 are solutions for the equation (we discard 0 because $0^0$ is indeterminated and $\frac{1}{0}$ is infinite). So we have one of the solutions, which is $\boxed{1}$ .

Now we can take into account the geometric series infinite sum when $|r|<1$ , which is $\frac{a_1}{1-r}$ , where $a_1$ is the first term of the geometric progession, which in this case is the same as r , $\frac{1}{x}$ :

$x^{\frac{\frac{1}{x}}{1-\frac{1}{x}}}=x^x\Rightarrow x^{\frac{\frac{1}{x}}{\frac{x-1}{x}}}=x^x\Rightarrow x^{\color{#D61F06}\frac{1}{x-1}}=x^{\color{#D61F06}x}\Rightarrow \frac{1}{x-1} = x\Rightarrow\ x^2-x-1=0\Rightarrow \boxed{\mathbf{x=\frac{1\pm\sqrt{5}}{2}}}$

And we discovered two more solutions. However, $\frac{1-\sqrt{5}}{2}\approx -0.61$ , so $\left|\frac{1}{x}\right |<1\Rightarrow |r|\ngtr 1$ .Therefore, we discard the negative solution since its absolute value cannot be smaller than 1 because the series will diverge. Now, we conclude that the two real solutions are 1 and $\frac{1+\sqrt{5}}{2}=\phi$

Then the sum of both solutions is $\boxed{1+\phi\approx 2.61803}$