# Ramanujan's Infinite Nested Roots

## Theorem

$3 = \sqrt {1 + 2 \sqrt {1 + 3 \sqrt { 1 + \cdots} } }$

## Proof

We have:

 $\ds 3$ $=$ $\ds \sqrt 9$ $\ds$ $=$ $\ds \sqrt {1 + 8}$ $\ds$ $=$ $\ds \sqrt {1 + 2 \sqrt {16} }$ $\ds$ $=$ $\ds \sqrt {1 + 2 \sqrt {1 + 15} }$ $\ds$ $=$ $\ds \sqrt {1 + 2 \sqrt {1 + 3 \sqrt {25} } }$ $\ds$ $=$ $\ds \sqrt {1 + 2 \sqrt {1 + 3 \sqrt {1 + 24} } }$ $\ds$ $=$ $\ds \sqrt {1 + 2 \sqrt {1 + 3 \sqrt {1 + 4 \sqrt {36} } } }$

In order for this sequence to continue, it needs to be shown that:

$n + 1 = \sqrt {1 + n \left({n + 2}\right)}$

Thus:

 $\ds \sqrt {1 + n \left({n + 2}\right)}$ $=$ $\ds \sqrt {1 + n^2 + 2 n}$ $\ds$ $=$ $\ds \sqrt {\left({n + 1}\right)^2}$ Square of Sum $\ds$ $=$ $\ds n + 1$

The result follows.

$\blacksquare$

## Source of Name

This entry was named for Srinivasa Ramanujan.