Let \((s_n)_{n = 0}^{\infty}\) be a sequence of the form \(s_0 = \sqrt{a}, s_1 = \sqrt{a + \sqrt{a}}, s_2 = \sqrt{a + \sqrt{a + \sqrt{a}}}, ...\)

As \(n\) approaches infinity, the sequence approaches a "continued square root", a number of the following form:

\(x = \sqrt{a + \sqrt{a + \sqrt{a + ...}}}\)

I'll leave it as an exercise to show that if \(a > 0\), then this continued square root always converges, and the result is the positive root of the equation \(a = x^2 - x\).

Let's take \(a = 2\) as an example. By the formula I just mentioned, we have \(\sqrt{2+\sqrt{2+\sqrt{2+...}}} = 2\).

One question I asked myself is, how fast do these continued square roots converge to \(2\)?

To put it more precisely, let's consider the sequence \(2 - \sqrt{2}, 2 - \sqrt{2 + \sqrt{2}}, 2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}\), and so on. The decimal approximations for these numbers are 0.5858, 0.1522, 0.03843, 0.0096305, etc. Do you notice the pattern? The numbers seem to be getting about \(4\) times smaller each time.

Let's make this more rigorous. In the original sequence, we have \(s_{n+1} = \sqrt{2 + s_n}\). Therefore, \[\lim_{n \to \infty} \dfrac{2 - s_{n+1}}{2 - s_n} =\lim_{n \to \infty} \dfrac{2 - \sqrt{2 + s_n}}{2 - s_n} = \lim_{n \to \infty} \dfrac{(2 - \sqrt{2 + s_n})(2 + \sqrt{2 + s_n})}{(2 - s_n)(2 + \sqrt{2 + s_n})}\]\[ = \lim_{n \to \infty} \dfrac{4 - (2 + s_n)}{(2 - s_n)(2 + \sqrt{2 + s_n})} = \lim_{n \to \infty} \dfrac{1}{2 + \sqrt{2 + s_n}} = \dfrac{1}{4}\]

So, the differences between the continued square roots and their limit (2) do, in fact, approach a geometric sequence with \(\dfrac{1}{4}\) as the common ratio. Hence, \(s_n\) can be asymptotically approximated by a function \(s_n \approx 2 - \dfrac{b}{4^n}\).

Now, the million dollar question: What is this constant \(b\)?

In other words, what is \(\displaystyle\lim_{n\to\infty} 4^n(2 - s_n)\)?

I won't tell you the answer in case you want to solve it on your own. If you succeed, I'd love to see how you solved it so feel free to share it in the comments. If you'd like to see my solution, you can find it here.

Now let's generalize the previous result for the real number \(a\). Let's replace \(2\) with \(x^2 - x\), so that our sequence looks like this: \[s_n = \left\{\sqrt{x^2 - x}, \sqrt{(x^2 - x) + \sqrt{(x^2 - x)}}, \sqrt{(x^2 - x) + \sqrt{(x^2 - x)+\sqrt{x^2 - x}}} ...\right\}\] You can show (using a similar proof to mine above) that \[\lim_{n \to \infty} \dfrac{x - s_{n+1}}{x - s_n} = \dfrac{1}{2x}\]

Hence, the generalized sequence can be approximated by the function \(s_n \approx x - \dfrac{L_x}{(2x)^n}\) where \(L_x\) is the following limit: \[L_x = \lim_{n \to \infty} (2x)^n (x - s_n)\]

Now, I have a conjecture for you. I haven't been able to prove/disprove it yet; see if you can:

Let \(s_n\) be a sequence defined recursively by \(s_0 = \sqrt{x^2 - x}\) and \(s_{n+1} = \sqrt{(x^2 - x) + s_n}\) for \(n \in \mathbb{N}\), for some \(x > 1\). Let \(L_x = \lim_{n \to \infty} (2x)^n (x - s_n)\). Then,

\[ \lim_{x \to \infty} L_x = \dfrac{1}{2}\]

## Comments

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TopNewestJust to confirm, is \(b=\dfrac{\pi^2}{16}\) ? – Ishan Singh · 1 year, 10 months ago

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– Ariel Gershon · 1 year, 10 months ago

Yes, very good!Log in to reply

– Ishan Singh · 1 year, 10 months ago

If we can solve the functional equation \(g(x)=a g^2\left(\dfrac{kx}{a}\right) - 1\) where \(a\) is a fixed constant and \(k\) is arbitarary, we are almost done solving the conjecture.Log in to reply

– Ariel Gershon · 1 year, 10 months ago

I'm sorry, how would this help?Log in to reply

– Ishan Singh · 1 year, 10 months ago

Let \(f_{n}(y)=\underbrace{\sqrt{a+\sqrt{a+\ldots+\sqrt{a+y}}}}_{n \ \text{times}}\), substitute \(y=a\cdot g(x)\) to simplify \(f_{n}(y)\). We can then replace \(a\) by \(t^2-t\) and calculate \(L_{t}\) by letting \(n \to \infty\). Then we can find the conjectured limit by letting \(t \to \infty\). For example if we let \(a=2\), \(k=1\) we will get \(g(x)=2g^2(\frac{x}{2})-1\). One solution is \(g(x)= \cos (x)\). From this we can calculate \(b=\dfrac{\pi^2}{16}\).Log in to reply

– Ariel Gershon · 1 year, 10 months ago

I'm missing quite a few steps in your logic.... What are \(g(x)\) and \(k\)?Log in to reply

– Ishan Singh · 1 year, 10 months ago

\(g(x)\) is an unknown function which satisfies \(g(x)=a g^2\left(\dfrac{kx}{a}\right) - 1\). If we substitute \(y=a\cdot g(x)\) in \(f_{n}(y)=\underbrace{\sqrt{a+\sqrt{a+\ldots+\sqrt{a+y}}}}_{n \ \text{times}}\), we have \(f_{n} (y) = a \cdot g\left(\dfrac{k^nx}{a^n}\right)\). And if we replace \(a\) by \(t^2-t\), we can then calculate \(L_{t}\) as a function of \(t\). This will simplify the conjectured limit \(L_{t}\). If my reasoning is missing any steps, you can ask and I'll try to fix it.Log in to reply

There are two more stipulations for \(g\) and \(c\) which you didn't mention, but they are important. The proof by induction of this approach requires that the following be true:

(1) there must exist a real number \(z\) such that \(g(z) = \dfrac{\sqrt{a}}{a}\)

(2) \(g(c^n z) \ge 0\) for all \(n \in \mathbb{N}\)

Also, we can write \(f_n(a) = s_n\). Then by induction we can show that \(s_n = a * g(c^n z)\). If we replace \(a\) with \(t^2 - t\), then \(L_t\) will have the following formula: \[L_t = \lim_{n\to\infty} (2t)^n \left(t - (t^2 - t) g(c^n z)\right)\]

In the case of \(t = 2\), we can have \(c = \dfrac{1}{2}\) and \(g(x) = \cos(x)\), since, as you mentioned, \(\cos(x) = 2\cos^2\left(\dfrac{x}{2}\right) - 1\). We have \(z = \dfrac{\pi}{4}\) since \(\cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}\). Then we get \[L_2 = \lim_{n\to\infty} 4^n \left(2 - 2\cos\left(\dfrac{\pi}{4*2^n}\right)\right)\] This limit does, in fact, evaluate to \(\dfrac{\pi^2}{16}\).

Interesting as this is, though, I think it won't be as easy to find \(g, c, z\) for \(t\) in general. – Ariel Gershon · 1 year, 10 months ago

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I want to point out that it only applies for \(a > 0\). (For \(a = 0\), it converges to the

lowerroot of \(a = x^2-x\), namely \(x = 0\) instead of \(x = 1\).) – Ivan Koswara · 1 year, 10 months agoLog in to reply

– Ariel Gershon · 1 year, 10 months ago

Fair point. I'll fix itLog in to reply