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We know that from using Ramanujan formula, we have :
$$3 = \sqrt{1+2\sqrt{1+\sqrt{3+\cdots}}}$$ Now, suppose I define the following sequence : $$ x_1 = 1\\ x_2 = \sqrt{1+\sqrt{2}}\\ \vdots\\ x_n = \sqrt{1+\sqrt{2+\sqrt{3+\cdots}}} $$ I have checked numerically the the value for $x_n$ is approximately $1.7$ but it does not equate to $\sqrt3$.

Can anyone help me to solve the exact value of this sequence?

Michael Hardy
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Evan William Chandra
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  • Notice that the sequence you defined does not quite resemble the expression from Ramanujan's formula. – Fimpellizzeri May 11 '17 at 17:38
  • Related: https://math.stackexchange.com/questions/774193/is-the-nested-radical-constant-rational-or-irrational – gt6989b May 11 '17 at 17:54
  • Also Related: https://math.stackexchange.com/questions/437209/how-can-i-show-that-sqrt1-sqrt2-sqrt3-sqrt-ldots-exists – gt6989b May 11 '17 at 17:54

2 Answers2

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The value to which the sequence converges is called the Nested radical constant, no closed-form expression is known for this constant.

kingW3
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Here is perhaps an approach. Let $$ A_k = \sqrt{k + \sqrt{k+1 + \sqrt{\ldots}}} $$ then assuming $(A_k)$ exists, $$ A_{k+1} = A_k^2-k $$ and $A_1$ will give you what you are seeking.

gt6989b
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