A calculus problem by A Former Brilliant Member

Calculus Level 2

$\large{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots }}}}} = \, ?$

3 solutions

Chew-Seong Cheong
Oct 8, 2017

Ramanujan's formula of nested radical (eqn. 26) is as follows:

$x+n+a = \sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{a(x+2n)+(n+a)^2+(x+2n)\sqrt \cdots}}}$

Putting $x=2$ , $n=1$ and $a=0$ , we have:

$2+1+0 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt\cdots}}} = \boxed{3}$

A Former Brilliant Member
Oct 6, 2017

Here, you started from the answer, and showed why it was correct. But how would you get the answer in the first place?

Siva Budaraju - 3 years, 8 months ago

How do you know it remains true forever?

Kenny O. - 3 years, 8 months ago

Vilakshan Gupta
Oct 7, 2017

This is a very famous problem originally posed by Ramanujan