## Nested Radicals Yield Cool Results

by

While browsing at Barnes & Noble, I came across Excursions in Number Theory by Ogilvy and Anderson. “Hey! I teach number theory,” I thought. So I flipped through the book for a while and came across the following question:

Let $n$ be an integer, and consider the expression

$\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}}$

Does this ever converge? If so, to what?

Assume the expression converges (what it means for something like that to converge is left for another post), and let $x$ be the limit. Then

$x^2-n=x,$

and $x = \frac{1}{2}(1+\sqrt{1+4n})$. In particular, the expression converges. What’s more, if $n=k(k-1)$ for some integer $k$, then

$x = \frac{1}{2}(1+\sqrt{1+4n}) = k,$

an integer! For example, with $k=2$, so that $n=2$, we have

$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}} = 2.$

Neat! Mathworld has a lot more on nested radicals. Especially cool are the expressions for $\pi$:

$\dfrac{2}{\pi} = \sqrt{\dfrac{1}{2}} \sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2}}} \sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{1}{2}}}}\cdots$

and the golden ration $\phi$:

$\phi = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$

### 4 Responses to “Nested Radicals Yield Cool Results”

1. heather360 Says:

Cool! I was able to see that it would be the Golden ratio and then an integer in some cases.

Nested roots and nested (continued) fractions have always seemed stranger to me than infinite series, which are pretty odd themselves.

2. TwoPi Says:

I have a nice proof that the nested radicals converge for positive n.

My proof does not generalize for negative n, although numerical experimentation has me convinced that the nested radicals still converge for that case as well, (of course) to the same expression (1/2 (1+sqrt(1+4n)).

3. Ξ Says:

In the complex case, wouldn’t this mean that any number 0.5 + i*(b) can be written as that nested root, for any b? And for real numbers, would it be the case that any real number x would be the root of nested ns for some real n (not necessarily an integer)? That’s kind of a weird distinction between real and complex numbers, if I’m envisioning this correctly.

4. TwoPi Says:

Yes I agree with your conclusions, provided we restrict ourselves to real values for n. (And I’m reasonably sure the expression actually converges for all real n as well.)

If we take convergence for granted, you can get any complex value as the limit of nested radicals if you allow n to be complex. But I’m not certain about convergence in this generalization.