3
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This question already has an answer here:

Kindly tell me what is the value of:

$\sqrt{1+2\times\sqrt{1+3\times\sqrt{1+4\times\sqrt\cdot.....}}}$

According to ramanujan ,it is equal to 3

I want to know how...

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marked as duplicate by Hans Lundmark, Alex M., Matthew Conroy, Ross Millikan calculus Dec 6 '16 at 22:11

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.

  • $\begingroup$ You should be using backward slashes, \, instead of forward slashes, /, to format your question. $\endgroup$ – RGS Dec 5 '16 at 11:14
  • $\begingroup$ In third line it is 3,,, not 3!=6,,, $\endgroup$ – Atul Mishra Dec 5 '16 at 11:14
  • $\begingroup$ What might help : If we replace the square root after the number $n$ with $n+2$ , we get exactly $3$, for example $$\sqrt{1+2\cdot \sqrt{1+3\cdot \sqrt{1+4\cdot 6}}}=3$$ $\endgroup$ – Peter Dec 5 '16 at 13:51
  • $\begingroup$ I've seen ramanujan's solutions but I am not satisfied that this will work up to infinity....? $\endgroup$ – Atul Mishra Dec 5 '16 at 14:53
1
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This formula makes no sense. When u write "$\ldots$" it's like saying "and so on" where you expect the reader to understand from context what you are talking about.

For example if commonly when u see $1+2+3+4+\ldots$ you would think about a limit of sequence of partial sums therefore you would said $1+2+3+4+\ldots=\infty$ since that sequence does not converge.

However, speaking of ramanujan, he used different context where he stated that $1+2+3+4+\ldots=-{1 \over 12}$ simply because he meant something different by "$\ldots$"

Returning to your question. Your "$\ldots$" does not point to any scheme. ramanujan wrote something like it's 3 because

$3=\sqrt{9}=\sqrt{1+2\cdot4}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\cdot5}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot6}}}=\ldots=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}}$

Unfortunately such argument would work for any positive value.

For example $10=\sqrt{100}=\sqrt{1+2\cdot{99 \over 2}}=\sqrt{1+2\sqrt{1+3\cdot{ 9797 \over 12}}}=\ldots$

the weird fraction will become irrelevant

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