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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?

Shane Dizzy Sukardy
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    There are much easier ways to "prove" $4=3$ by essentially the same reasoning. For instance, $4=3+1=3+0+1=3+0+0+1=3+0+0+0+1=\dots=3+0+0+\dots=3$. Can you spot the error there? – Eric Wofsey Oct 06 '19 at 05:17

3 Answers3

3

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.

Nilotpal Sinha
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2

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;

$$3+1=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6...}}}}$$

user712576
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0

The generating function is

$$ f(x) = \sqrt{1+(x+1)f(x+1)}\Rightarrow f^2(x) = 1+(x+1)f(x+1) $$

so making $f(x) = a x + b$ we get $a = 1, b= 2$ or

$$ f(x) = x + 2 $$

Cesareo
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