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My Attempt: I tried to use the regular way.



But then I saw that nothing happened twice and I stopped. Any hints?

It's better to don't use limits but if you used no problem. If there is an answer that doesn't use limits will accept.

update1: The question that look likes this wants to find the limit but I want a much easier way to solve it. But if there isn't any answer easier answer the limition one will accept.

update2: You solve the problem when you know the answer is 3 think that you don't know that the answer is 3 then solve.


marked as duplicate by Yuriy S, Watson, Jack D'Aurizio calculus Jun 17 '16 at 14:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is a Ramanujan identity... I believe the answer to be $2$ $\endgroup$ – Frank Jun 17 '16 at 13:31
  • $\begingroup$ Can you tell your sloution please? $\endgroup$ – Taha Akbari Jun 17 '16 at 13:31
  • 2
    $\begingroup$ Wikipedia describes a simple procedure... using a formula that Ramanujan found: $\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(x+a)^2+(x+n)^2\sqrt{\ldots}}}$ $\endgroup$ – Frank Jun 17 '16 at 13:34
  • $\begingroup$ Can you link please? $\endgroup$ – Taha Akbari Jun 17 '16 at 13:34
  • 3
    $\begingroup$ en.wikipedia.org/wiki/Nested_radical under the tab Infinitely Nested Radicals. $\endgroup$ – Frank Jun 17 '16 at 13:34

Setting the Scene

Evaluate $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }}$

Let $$f_n(0)=\sqrt{1+n}$$also that $$f_n(k)=\sqrt{1+(n-k)f_n(k-1)}.$$

Then clearly $0<f_n(0)<n+1$ when $n>0$. Now if we assume that $f_n(k)<n+1-k$ and by induction that we see that $$ f_n(k+1) < \sqrt{1+(n-k-1)(n-k+1)} = \sqrt{1+(n-k)^2-1} = n+1-(k+1) $$ for all k. The expression is $f_n(n-2)$ which is increasing in $n$ and bounded above by $3$ and thus converges. Hence

$\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times }}=3$


Showing that $3$ is the limit can be achieved by writing

\begin{align} 3 &= \sqrt{9} \\ &= \sqrt{1 + 8}\\ &= \sqrt{1 + 2 \cdot 4} \\ &= \sqrt{1 + 2\sqrt{16}}\\ &= \sqrt{1 + 2\sqrt{1 + 3 \cdot 5}}\\ & = \sqrt{1 + 2\sqrt{1 + 3 \sqrt{25}}} \\ &= \sqrt{1 + 2\sqrt{1 + 3 \sqrt{1 + 4 \cdot 6}}}\\ &= \ldots \end{align} and so on.

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    $\begingroup$ You haven't shown the limit, you just proved that it converges. It must be shown that the supremum is 3, then the limit is also 3. With the same reasoning I could say the limit is 69, because it is an upper bound. $\endgroup$ – user305860 Jun 17 '16 at 13:51
  • $\begingroup$ How about this: $$4=\sqrt{1+2\cdot \frac {15}{2}}=\sqrt{1+2\sqrt{1+3\cdot \frac {221}{12}}}=\sqrt{1+2\sqrt{1+3\sqrt{1+\ldots}}}$$Thus, the infinitely nested radical is equal to $4$ too! (A contradiction) $\endgroup$ – Frank Oct 19 '16 at 1:53

First you can guess that this seems to converge to $3$ by using a calculator (or by estimating the radicals). Assuming the limit is three, then one can try to wrtie $3$ as the following down:

$3=\sqrt{1+8}=\sqrt{1+2\cdot4}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+15}}=\sqrt{1+2\sqrt{1+3*5}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot 6}}}=\ldots$

This iteration is the same as the formula you stated above, because it holds

$\sqrt{n^2}=\sqrt{1+(n-1)(n+1)}=\sqrt{1+(n-1)\sqrt{(n+1)^2}} \quad \forall n\in \mathbb{N}, n>1$

Thus, the limit is $3$.


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