# More infinite nested square roots

Let us consider the following nested square root: $$A=\sqrt{b_1+\sqrt{b_2+ \sqrt{b_3+\cdots}} }.$$

Consider the following three cases:

• $$b_k = k^2 + k -1 \Rightarrow A = 2.$$
• $$b_k = (k+2)^4 - 4(k+2)^3+5(k+2)^2-4(k+2)+1 \Rightarrow A = 4.$$
• $$b_k = 4^{k-1} \Rightarrow A = 2.$$

The last case (the methodology) is discussed in my previous MSE question, here. These are very particular cases of a methodology that can create many similar formulas for nested square roots, nested cubic roots, or continued fractions. The first formula could be related to the Ramanujan's infinite radicals (see here) though I am not sure - this Wikipedia article is quite confusing. Also the third case is not new, indeed it validates my approach.

This does not seem to be a very advanced topic, though convergence issues are not simple in some cases. For instance, the case $$b_1 = 1, b_2 = 7, b_{k+1} = 3b_k(b_k -1)$$ if $$k>2$$, using nested cubic roots rather than nested square roots, does converge but not to the simple value predicted by my scheme.

Before going any deeper into this, I'd like to have people look into the first two formulas and try to prove the result. The first two cases involve an algorithm with $$b_k = a_{k+2}, c_{k+1}= (c_k - a_k)^2$$ if $$k>2$$, $$c_2 = 4, a_k = \lfloor c_k - k^\alpha \rfloor$$ if $$k>1$$, and $$a_1 =1$$. The first case corresponds to $$\alpha =1$$, the second case to $$\alpha = 2$$.

For the first case, we can start with $$\:2 = \sqrt{1+\sqrt{9}} = \sqrt{(1^2+1-1)+\sqrt{(2+1)^2}}\:$$. Then we repeatedly apply the identity

$$(k+1)^2 = (k^2+k-1) + (k+2) = (k^2+k-1) + \sqrt{(k+2)^2}$$

So we get

\begin{align} 2 = \sqrt{1+\sqrt{9}} &= \sqrt{(1^2+1-1)+\sqrt{(2+1)^2}}\\[1.5ex] &= \sqrt{(1^2+1-1)+\sqrt{2^2+2-1 + \sqrt{(3+1)^2}}}\\[1.5ex] &= \sqrt{(1^2+1-1)+\sqrt{2^2+2-1 + \sqrt{3^2+3-1 +\sqrt{(4+1)^2}}}}\\[1.5ex] &= \; ... \end{align}

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In the second case, the identity $$\:(x-1)^4 = (x^4-4x^3+5x^2-4x+1) + x^2\:$$ leads to

\begin{align} \color{white}{text}\\ (k+1)^4 &= \left[(k+2)^4-4(k+2)^3+5(k+2)^2-4(k+2)+1\right] + (k+2)^2\\[1.5ex] &= b_k + \sqrt{\left[(k+1)+1\right]^4}\\ \color{white}{text}\\ \end{align}

Then, starting with $$\:4 = \sqrt{7+\sqrt{81}} = \sqrt{b_1+\sqrt{(2+1)^4}}\:$$, we repeatedly apply the above identity and get

\begin{align} 4 &= \sqrt{7+\sqrt{81}}\\[1.5ex] &= \sqrt{b_1+\sqrt{(2+1)^4}}\\[1.5ex] &= \sqrt{b_1+\sqrt{b_2 + \sqrt{(3+1)^4}}}\\[1.5ex] &= \sqrt{b_1+\sqrt{b_2+\sqrt{b_3 + \sqrt{(4+1)^4}}}}\\[1.5ex] &= \; ... \end{align}

$$\color{white}{text}$$

[P.S. Would appreciate any suggestions to make those last lines in each proof look better; or is three the highest number of nested radicals that MathJax can handle well?]