## July 2017

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My July diary included the following brainteaser.

Does √(1+2√(1+3√(1+4√(1+5√(1+ …))))) have a finite numerical value? If so, what is it?

The expression in question looks neater if the square roots are written with overlines. Unfortunately that is beyond my HTML-coding capabilities, but here's an image:

I described this in the diary as "an old chestnut." For sure it's at least 106 years old. It was posed by
Srinivasa Ramanujan as Question 289 in Volume 3 of the *Journal of the Indian
Mathematical Society* (1911). The background to that, although not the solution I give below, is in Chapter Three of
Robert Kanigel's 1991 book about Ramanujan.

It's a cute problem, under the heading "deceptively simple." People not familiar with it get into all sorts of trouble. I've seen them resorting to advanced calculus.

Kanigel writes:

Three issues of theJournalcame and went — six months — with no solution offered; in the end, Ramanujan supplied it himself.

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**• Solution**

I shall give boldface Roman numbers and sub-numbers to all my key statements here.

The first statement is true for any positive real number *a*, by definition:

I.ia= √(a²)

Rewriting that slightly:

I.iia= √(1 + (a² − 1))

Using a well-known bit of junior-high algebra, I can rewrite *that* as:

I.iiia= √(1 + (a− 1)(a+ 1))

Since that is true for any number *a*, it's just as true for *a* + 1. So, substituting *a* + 1 for
*a* in **I.iii**:

IIa+ 1 = √(1 +a(a+ 2))

**I.iii** is also true for *a* + 2, of course. So, substituting *a* + 2 for
*a* in **I.iii**:

IIIa+ 2 = √(1 + (a+ 1)(a+ 3))

This also works for *a* + 3 …

IVa+ 3 = √(1 + (a+ 2)(a+ 4))

… for *a* + 4 …

Va+ 4 = √(1 + (a+ 3)(a+ 5))

… and so on.

Now substitute the value of *a* + 1 given by **II** back into **I.iii**:

I.iva= √(1 + (a− 1)√(1 +a(a+ 2)))

Substitute the value of *a* + 2 given by **III** back into *that* to get **I.v**; substitute the value of
*a* + 3 given by **IV** into **I.v** to get **I.vi**, and … keep going.

You end up with an infinitely long expression which, if you set *a* = 3, is the one to be evaluated. The answer is therefore
*a* = 3.

As I said, deceptively simple. The hardest math here is that bit of junior-high algebra.

No, that's not right. The hardest math here is *seeing your way to a simple and elegant solution*. That is *genius*-level math.