While I was surfing on the Internet yesterday, I watched a video about Ramajuan's infinite root. After that I had tried on my own and I got the point.

$ 3=\sqrt9$
$3=\sqrt{1+2 \cdot 4}$
$3=\sqrt{1+2\cdot \sqrt{16}}$
$3=\sqrt{1+2\cdot \sqrt{1+15}}$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot 5} }$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{25}} }$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+24}} }$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot 6}} }$
$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{36}}} }$

$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$

and I also tried

$ 4=\sqrt{16}$
$4=\sqrt{1+2 \cdot \frac{15}{2}}$
$4=\sqrt{1+2\cdot \sqrt{\frac{225}{4}}}$
$4=\sqrt{1+2\cdot \sqrt{1+\frac{221}{4}}}$
$4=\sqrt{1+2\cdot \sqrt{1+ 3\cdot \frac{221}{12}}}$
$4=\sqrt{1+2\cdot \sqrt{1+ 3\cdot \sqrt{\frac{48841}{144}}}}$
$4=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$

for the value 4. Isn't there a contradiction? I also tried with starting $2=\sqrt4$ and it does not work for infinetely times . I sensed that I can do same actions for every number which is greater than and equals to 3. For every help and opinion, thanks in advance :)

  • 1
    $\begingroup$ Very much reminds me of this. $\endgroup$ Aug 3, 2017 at 12:39
  • 1
    $\begingroup$ When you are trying to continue some process to infinity, it's not always going to work. There's a property called "convergence" $\endgroup$
    – Yuriy S
    Aug 3, 2017 at 12:47

1 Answer 1


There's no contradiction, there's just ill-defined terms.

The row

$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt36}} }$

is correct, but the following


$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$

is meaningless.

There is no clear way to define what the three dots at the end of that expression mean.

To provide a counterexample, when a mathematitian writes $\dots$, he is usually capable of defining exactly what those dots mean. For example, writing

$$1+\frac12+\frac 14 + \cdots$$

is the same as

$$\lim_{n\to\infty} 1+\frac12+\frac14+\cdots + \frac{1}{2^n}.$$

This is a well defined expression, because

  1. The concept of limit is well defined using $\epsilon-\delta$ definitions
  2. The concept of a finite sum of $n$ elements is defined through induction

If you want to go down deeper, every concept can be broken down into small pieces until only axioms and definitions remain.

On the other hand, there is no clear way how to strictly mathematically (using more basic concepts) write down what

$$\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$$ represents, and until you write it down strictly, you shouldn't be surprised when you get "weird" results. But those weird results aren't strictly speaking "contradictions", since they are not real mathematical expressions.

Another example of why you can't just write down three dots and hope everything sticks in infinity:

$$0.9\neq 1 \\0.99\neq 1 \\0.999\neq 1 \\\vdots \\0.999999\dots \neq 1$$

is also not true, because the conclusion is false.

Or, even more simply

  • $\{1\}$ is finite
  • $\{1,2\}$ is finite
  • $\{1,2,3\}$ is finite
  • $\vdots$
  • $\{1,2,3,\dots\}$ is finite.
  • $\begingroup$ Let$$f(n,k)=\begin{cases}\sqrt{1+kf(n,k+1)},&k<n\\n+1,&k=n\end{cases}$$Then you could interpret that infinitely nested radical as$$\lim_{n\to\infty}f(n,2)=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\dots}}}}$$More or less what Wojowu said. For this specifically, its identically $3$. $\endgroup$ Aug 3, 2017 at 12:52
  • 1
    $\begingroup$ There is a clear way to make sense out of Ramanujan's radical - let it be the limit of $\sqrt{1+2\cdot\sqrt{1+3\cdot\sqrt{\dots+n\cdot\sqrt{1}}}}$. $\endgroup$
    – Wojowu
    Aug 3, 2017 at 12:52
  • $\begingroup$ Thanks for your good and understandable examples and counterexamples. It's convergency is more clear now :) thanks for your time $\endgroup$
    – Alper
    Aug 3, 2017 at 13:15
  • $\begingroup$ @Wojowu is there a typo there? Why write $\sqrt{1}$ ??? $\endgroup$ Aug 3, 2017 at 14:48
  • $\begingroup$ @BrevanEllefsen Because this is exactly what you get if you remove $+\dots$ from the radical. $\endgroup$
    – Wojowu
    Aug 3, 2017 at 14:52

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