<s>[INST] Ramanujan's Nested Radical By noting Ramanujan's Nested Radical, we have
$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$
On the other hand, we can manipulate the number $4$ by applying the similar principle. Here we have
$\begin{aligned} 4 & = \sqrt{16} \\ & = \sqrt{1+15} \\ & = \sqrt{1+2 \cdot \dfrac{15}{2}} \\ & = \sqrt{1+2\sqrt{\dfrac{225}{4}}} \\ & = \sqrt{1+2\sqrt{1+\dfrac{221}{4}}} \\ & = \sqrt{1+2\sqrt{1+3 \cdot \dfrac{221}{12}}} \\ & = \sqrt{1+2\sqrt{1+3\sqrt{\dfrac{44841}{144}}}} \\ & \vdots \\ & = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}} \end{aligned}$
How can it be? Something contradicts?
 [/INST] By Ramanujan's nested radical, we can also get this curious identity;
$$n+m=\sqrt{m^2+n\sqrt{m^2+(n+m)\sqrt{m^2+(n+2m)\sqrt{..}}}}$$
Ramanujan proved this identity, and if you want a look at how it's done in a simple manner (not-rigorous), check out this blog. And by plugging in $4$, we get;
$$3+1=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6...}}}}$$
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