The infinite radicals of Ramanujan

What if I asked you to find the numerical value of

\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{\cdots}}}}}} ?

If you have difficulty figuring out what this thing is, don’t worry. You’re not the only one. This question is one the problems posed by Srinivasa Ramanujan, one of the most brilliant and, well, mysterious mathematicians of all time. In fact, it was head enough that Ramanujan had to give the answer a few months later.

So, what is the value of this infinite series of nested radicals?

If we try figuring the limit using progressively more terms, we get:







and if we continue long enough, we see that we inch ever closer to 3. So is:

\displaystyle 3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{\cdots}}}}}} ?

Yes! Let’s see how we can arrive at this result:











continuing long enough,


* *

Is this method applicable to any n, and not just 3? That’s a legitimate question because it might be so that 3 is
just some special case. Well, turns out that, no, if n is a natural
number, it always works:



and since n^2-1 can always be factored as n^2-1=(n-1)(n+1),




and (n+1)^2-1 being of the general form $m^2-1=(m-1)(m+1)$, we can factor it again and continue the expansion.

* *

This method is related to the Court of Lions method I presented a good while ago to compute square roots. This method, based on a simple geometric operation, allows us to write

\displaystyle \sqrt{5}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1}^2}^2}^2}^2}.

This allows us to build a line segment of length \sqrt{n} for arbitrary n using only a ruler an a compass. Not use how useful it really is, though.

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