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
- 1989: Bruce C. Berndt: Ramanujan's Notebooks: Part II: Chapter $\text {12}$. Continued Fractions
- Feb. 1986: Walter S. Sizer: Continued Roots (Math. Mag. Vol. 49, no. 1: pp. 23 – 27) www.jstor.org/stable/2690013