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## Infinitely nested radicals posed by Ramanujan

### Solution

Solution by the mathematician Srinivasa Ramanujan :
$n(n+2)=n\sqrt{1+(n+1)(n+3)}$ Let $$n(n+2)=f(n)$$
Thus, we see that \begin{aligned} f(n) &=n \sqrt{1+f(n+1)} \\\\ &=n \sqrt{1+(n+1)\sqrt{1+f(n+2)}}\\\\ &=n \sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+f(n+3)}}}\\\\ &=n \sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+\cdot\cdot\cdot}}}} \end{aligned}
$\Rightarrow n(n+2)=n \sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+\cdot\cdot\cdot}}}}$
Putting $$n=1$$, we get $\large \sqrt{1+2\sqrt{1+3\sqrt{1+\cdot\cdot\cdot}}}=3$

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