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Let $$x_n=\sqrt{1+\sqrt{2+\sqrt{3+\dots+\sqrt{n}}}}$$ Show that $x_n<x_{n+1}$ and $x_{n+1}^2\le 1+\sqrt{2}x_n$. Hence show that $x_n$ is bounded by $2$ and converges.

Trials:I can show that $x_n<x_{n+1}$ and $$x_{n+1}^2= 1+\sqrt{2+\sqrt{3+\dots+\sqrt{n}}}\\=1+\sqrt 2[1+\sqrt{3/4+\dots+\sqrt{n/2^{n-1}}}]\ \\\le1+\sqrt 2 x_n$$ Then I can't show $x_n$ is bounded by $2$ and converges. Please help.

Souvik Naskar
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3 Answers3

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Hint: $x_n$ increases, $x_1 < 2$ and $1+2\sqrt2 < 2^2$.

njguliyev
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Two hints:

First, you get $x_{n+1}$ by replacing the innermost $\sqrt{n}$ in the expression for $x_n$ by $\sqrt{n+\sqrt{n+1}}$, which is larger.

Second, from the two inequalities proved you get $x_{n+1}^2\le1+\sqrt2x_n<1+\sqrt2x_{n+1}$.

Added in response to a question in the comments: Now complete the square (write just $x$ instead of $x_{n+1}$). You get, in order: $$\begin{gather*} x^2-\sqrt2x+\frac12<\frac32 \\ \Bigl(x-\frac1{\sqrt2}\Bigr)^2<\frac32 \\ x<\frac{1+\sqrt3}{\sqrt2} \end{gather*} $$ which is actually a better estimate than required (the upper bound is $1.9318…$).

Harald Hanche-Olsen
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$x_1=1\le 2$. Suppose $x_n\le 2$, then $x_{n+1}^2\le 1+\sqrt 2x_n\le 1+2\sqrt 2$. We have $8\le 9 \Rightarrow 2\sqrt 2\le 3 \Rightarrow 1+2\sqrt 2\le 4$. Therefore, $x_{n+1}\le 2$ and by induction this holds for all $n$. Now $x_n$ is a bounded increasing sequence and hence converges.

walcher
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