Nested Square Roots by Ramanujan

The Indian mathematician Srinivasa Ramanujan was as an extremely talented autodidact, capable of doing the most complicated algebraic manipulations. He later worked with Hardy in England fighting with lots of health problems there. He died after he returned to Madras in 1920 at the age of 32. Ramanujan invented many admirable formulas, many of which were later used by other mathematicians.

3 = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{ 1 + 4 \sqrt{ \ldots }}}} 

This is a relatively simple one. From a modern viewpoint, the presentation shown above lacks some precision. As we will see, it is by far not as clear as we would want it to be. Mathematics is also about clarity of thought and writing. You find this formula in a lot of articles, and nowadays in YouTube videos. Nothing that I can say about it might be new.

Let us try to make sense of the formula. It is based on the simple observation that

3 = \sqrt{1 + 2 \cdot 4}, \quad 4 = \sqrt{1 + 3 \cdot 5}, \quad 5 = \sqrt{1 + 4 \cdot 6}, \quad \ldots

or in general

n = \sqrt{1 + (n-1) \cdot (n+1)}.

That is an easy application of one of the Binomial formulas. If you stack this together, you end with the nested square roots

3 = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{ 1 + 4 \sqrt{ 1 + \ldots \sqrt{1 + n \sqrt{ R_n }}}}}}

which is correct for all n, if you insert

R_n = (n+2)^2

So, the meaning of the „…“ part in Ramanujan’s equation is an ever growing sequence of natural numbers. Ramanujan did not think of it that way. He just put his stacked insertions into a simple question:

\text{What is \quad} \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{ 1 + 4 \sqrt{ \ldots }}}} \text{\quad ?}

He posed this into a newspaper as a problem and got no answers. Let us approach this question in a more modern, systematic way.

We often think of „…“ as a process which repeats in a systematic algorithmic way, as in

2 = 1 + \frac12 + (\frac12)^2 + (\frac12)^3 + (\frac12)^4 + \ldots

We can truncate the expression on the right in front of „+ …“ at any point and get a converging series. Or, if we write it as above with residuals R(n)

2 = 1 + \frac12 + (\frac12)^2 + (\frac12)^3 + \ldots + (\frac12)^n + R_n \qquad \text{with } R_n \to 0.

This is not the understanding of Ramanujan. His „…“ do not converge to 0. In fact, he never considered convergence in his early work at all, as far as I know.

We are talking about algebraic identities here. So we understand the question in a way such that the value does not change during the process of making the „…“ longer, i.e.,

\sqrt{1 + \ldots \sqrt{1 + n \sqrt{ R_n }}} = \sqrt{1 + \ldots \sqrt{1 + n \sqrt{ 1 + (n+1)\sqrt{R_{n+1} }}}}.

For R(n)>=1 this is equivalent to the recursion formula

R_{n+1} = \left(\frac{R_n-1}{n+1}\right)^2 \tag{*}

If R(n)<1, the process breaks down, and we are unable to satisfy the identity. We can now start with any big enough value for R(1) and get the desired identity. Ramanujan’s choice was R(1) = 9 which yields value 3

3 = \sqrt{R_1} = \sqrt{9}, \quad 3 = \sqrt{1 + 2\sqrt{R_2}} = \sqrt{1 + 2 \sqrt{16}}, \quad \ldots

with the nice sequence

R_n = (n+2)^2.

This start choice, however, is arbitrary. We can take any R(1)>9 too and get another sequence. The R(n) will not look as pleasing, but it works.

If the readers of the newspaper where the problem appeared were mathematically educated in a modern way, they should indeed find that the sequence can have many positive values, just by taking the right starting point. To formulate the problem for the ones that like algebraic manipulators, the value of Ramanujan’s equation depends on the choice of

R_1 = 1 + 2 \sqrt{1 + 3 \sqrt{ 1 + 4 \sqrt{ 1 + 5 \sqrt{ \ldots}}}} \quad \ge \quad 9

which actually means, it depends on the choice of the solution itself. This does not make much sense.

As usual, we can now try to start making interesting mathematics asking interesting questions. For instance, for which starting points does the sequences of residuals according to (*) stay above 1?

It turns out that any starting point <9 yields convergence to 0 in our recursion (*). So it would not work. And with any starting point greater or equal 9 the sequence goes to infinity. So it works. Thus, you can call Ramanujan’s choice a sensible one. The somewhat involved proof is left to the reader.

There is another interesting question. If we try to compute the value of this „infinite expression“ from inside out, we would have to associate a value with „…“ and compute ever longer expressions. Assume we take the same value for „…“. Then we get a sequence of results. In Euler Math Toolbox, we can simulate this as follows:

function f(n,a) ...
  for k=n to 2 step -1;
    a = sqrt(a)*k+1
  end;
  return sqrt(a)
endfunction

It turns out, that this process converges to 3 for any a. From this viewpoint, Ramanujan was right!

This result is connected to the fact that the maximal value for R(1) such that our recursion (*) is bounded is 3. For any value of R(n), you can select a starting point, and if you fix the values of R(n) to one specific value, this starting point will converge to 3. The proof is again a bit involved. If someone finds a nice paper about this, please write in the comments.