Ramanujan’s infinite radicals

Here’s one of my favorite puzzles.

The Problem

Solve the following equation:

Screen Shot 2018-07-28 at 7.00.13 AM


This problem was initially posed by the most brilliant mathematician of the past century, an eccentric prodigy named Srinivasa Ramanujan. He submitted it to the Journal of Indian Mathematical Society without a solution, challenging his fellow mathematicians to solve it. Eventually, when nobody could do it, he gave in and revealed the answer along with a simple proof that probably put them all to shame.


Ok, so seriously, try this for yourself before reading on. Most people haven’t ever done anything with infinite nested square roots, so this is a good opportunity to be creative and come up with weird ways of solving the problem. Play around with it, give random things names, and see what you can do!

You’ll notice (as I did when I first tried working it out) that solving nested radicals can be really really hard. Things quickly get really ugly. Luckily, the answer is beautifully simple. It’s just 3!

Let’s see why. We start with a very simple identity:

Screen Shot 2018-07-28 at 7.20.54 AM

Taking the square root, we get

Screen Shot 2018-07-28 at 7.21.02 AM

If we just plug in a few integer values, we see the following:

Screen Shot 2018-07-28 at 7.16.28 AM.png

What happens if we plug in the value of 4 from the second line to the 4 in the first? We get

Screen Shot 2018-07-28 at 7.18.10 AM.png

Hm, already looking familiar! We’re not done yet – we now plug in the value of 5 from the third line to this new equation, obtaining:

Screen Shot 2018-07-28 at 7.19.37 AM.png

And once more:

Screen Shot 2018-07-28 at 7.22.31 AM.png

And etc off to infinity.

Screen Shot 2018-07-28 at 7.42.25 AM.png

And that’s the proof!


Now, Ramanujan actually proved a much more general result than this. I’ll prove a less general result than his, but still more general than what we just saw.

The generalization comes from noticing that our initial identity didn’t have to involve adding 1 in (x+1)2. We could have added any number to x, squared it, and then gotten a brand new general theorem about a whole class of continued root problems.

Thus we start with the identity

Screen Shot 2018-07-28 at 7.30.03 AM


Screen Shot 2018-07-28 at 7.30.23 AM.png

Now we perform substitutions exactly as we did above:


In the end, what we get is a solution for a whole class of initially very difficult seeming problems:

Screen Shot 2018-07-28 at 7.39.58 AM.png

And if we plug in m = 1 and n = 2, we get our initial problem!

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