Ramanujan's Infinite Nested Roots

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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.


Sources