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?

In Ramanujan's radical, if you stop after $n$ nested radicals the last term inside the radical will be $\sqrt{1}$ but in your case, if you stop after $n$ nested radicals the last term is a term which is increasing with $n$. Hence the difference.
