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Problem

Evaluate $$x_n=\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots}}}.$$

My Reaserch

In fact, the problem requires to evaluate the limit of the sequence below as $n \to \infty$ $$x_n=\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots+(n-2)\sqrt{1+(n-1)\sqrt{1+n}}}}},$$where $n=1,2,\cdots.$

We may show that the limit really exists by $\textbf{Monotone Convergence Theorem}$. First, it's clear that $x_n$ is increasing with the increasing $n$. Second, we may obtain$$\begin{align*}x_n&<\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots+\sqrt{1+n\sqrt{(n+2)^2}}}}}\\&<\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots+\sqrt{1+(n-1)(n+1)}}}}\\&=\cdots\\&=\sqrt{1+2 \times 4}\\&=3.\end{align*}$$ Hitherto,we have showed that $x_n$ is increasing but with an upper bound $3$. Therefore, there exists a limit for $x_n$ as $n \to \infty$.

But how to go on with this? I'm stuck here.

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  • https://math.stackexchange.com/questions/7204/evaluating-the-nested-radical-sqrt1-2-sqrt1-3-sqrt1-cdots – Alex R. May 19 '18 at 06:21
  • Notice the following equalities: $$\begin{align} 3&=\sqrt{9} = \sqrt{1+8} = \sqrt{1+2\cdot 4} = \sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+15}} \ &=\sqrt{1+2\sqrt{1+3\cdot 5}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=\sqrt{1+2\sqrt{1+3\sqrt{1+24}}} \ &=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot 6}}} =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{36}}}}=\cdots\end{align}$$ $$\therefore x_n\stackrel{n\to \infty}{\longrightarrow}3$$ – Mr Pie May 19 '18 at 08:55

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