Addition never end

For computing nested fractions $\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{...}}}}}}}}}$ . Don't forget the quadratic equation.

First.

$\begin{aligned} R = \sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{...}}}}}}} &\Rightarrow R = \sqrt{C+x} \\&\Rightarrow x^2-x-C = 0 \\&\Rightarrow x = \dfrac{b+\sqrt{b^2-4a}}{2} \end{aligned}$

Go to nested functions for full knowledge about this.

Try it if it is converge.

Let

$\begin{aligned}x=&\ \sqrt{110+\sqrt{110+\sqrt{110+\sqrt{110+\cdots}}}}\\x^2=&\ 110+\sqrt{110+\sqrt{110+\sqrt{110+\cdots}}}\\x^2=&\ 110+x\\x^2-x-110=&\ 0\\(x-11)(x+10)=&\ 0\\x=&\ 11,-10\end{aligned}$

But $x$ should be positive. Thus, $\sqrt{110+\sqrt{110+\sqrt{110+\sqrt{110+\cdots}}}}=\boxed{11}$

so first of all, the answer is positive. We're not subtracting, as that would be a different story. So here, let's set the infinitely nested function as x. so we multiply both sides by x, we get 110 + x = $x^{2}$ . do some simple subtraction, then you get the equation; $x^{2}$ - x - 110 = 0. My way is factoring, where you split $x^{2}$ into $x \times x$ , then split 110 into -11 and +10 (since this is a negative number, $negative \times positive$ = negative), cross multiply, you get 10x - 11x = -x, exactly what we want. So now we have the equation (x + 10)(x - 11) = 0. Either of the brackets has to be 0, and there is no value that could substitute both x's, so in the end we have x = -10 or x = 11. ditch the negative answer a.k.a. -10, and you get 11.