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.
Three issues of the Journal came and went — six months — with no solution offered; in the end, Ramanujan supplied it himself.
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.i a = √(a²)
Rewriting that slightly:
I.ii a = √(1 + (a² − 1))
Using a well-known bit of junior-high algebra, I can rewrite that as:
I.iii a = √(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:
II a + 1 = √(1 + a(a + 2))
I.iii is also true for a + 2, of course. So, substituting a + 2 for a in I.iii:
III a + 2 = √(1 + (a + 1)(a + 3))
This also works for a + 3 …
IV a + 3 = √(1 + (a + 2)(a + 4))
… for a + 4 …
V a + 4 = √(1 + (a + 3)(a + 5))
… and so on.
Now substitute the value of a + 1 given by II back into I.iii:
I.iv a = √(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.