Showing that $\lim\limits_{n\to\infty}x_n$ exists, where $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + …+\sqrt{n}}}}$

Let $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$

a) Show that $x_{n} < x_{n+1}$

b) Show that $x_{n+1}^{2} \leq 1+ \sqrt{2} x_{n}$

Hint : Square $x_{n+1}$ and factor a 2 out of the square root

c) Hence Show that $x_{n}$ is bounded above by 2. Deduce that $\lim\limits_{n\to \infty} x_{n}$ exists.

Any help? I don't know where to start.

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You can't do part a? – wj32 Jun 25 '13 at 6:26

3 Answers

a) Note that $0<u<v$ implies $0<\sqrt u<\sqrt v$. This allows you to show the claim by starting from $0<n<n+\sqrt {n+1}$ and walking your way to the outer $\sqrt{}$.

b) Follow the hint

c) By induction: $0<x_1<2$ and $0<x_n<2$ implies $1+\sqrt 2 x_n<1+2\sqrt 2<4$

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Hints: Induction on

$$\bullet\;\;x_n<x_{n+1}\iff 1+\sqrt{2+\sqrt{3\ldots+\sqrt n}}<1+\sqrt{2+\sqrt{3+\ldots+\sqrt{n+\sqrt{n+1}}}}\iff$$

$$2+\sqrt{3+\ldots+\sqrt n}<2+\sqrt{3+\ldots\sqrt{n+1}}\iff\ldots$$

$$\bullet\bullet\;x_{n+1}^2=1+\sqrt{2+\sqrt{3+\ldots+\sqrt{n+1}}}\le 1+\sqrt2\left(\sqrt{1+\sqrt{2+\ldots+\sqrt n}}\right)=1+\sqrt2\,x_n$$

$$\iff\left(\sqrt{2+\sqrt{3+\ldots+\sqrt{n+1}}}\right)\le\sqrt{2+2\sqrt{2+\ldots+\sqrt n}}\iff\ldots$$

For (c) you're already done with (a)-(b) since then you have a monotone ascending sequence bounded from above, so the sequence's limit equals its supremum...

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10 days old question, but .

a) Is already clear, that $\sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} < \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n+1}}}}$ , because $\sqrt{n} <\sqrt{n} + \sqrt{n+1}$ which is trivial.
My point here is to give some opinion about b) and c), for me it's better to do the c) first. We know that : $$\sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} < \sqrt{p+\sqrt{p+\sqrt{p+ ... }}}$$ But it is only true for $q\leq p<\infty$ for $q \in \mathbb{Z}^{+}$. Because it is trivial that $$\sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} > \sqrt{1+\sqrt{1+\sqrt{1+ ... }}}$$ Let $x=\sqrt{2+\sqrt{2+\sqrt{2+ ... }}}$, then $x^2=2+ \sqrt{2+\sqrt{2+\sqrt{2+ ... }}} \rightarrow x^2-x-2=0$, thus $x=2$, because $x>0$. Now let's probe this equation : $$\sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}} \leq \sqrt{2+\sqrt{2+\sqrt{2+ ... }}}=2 \tag{1}$$ 2 is bigger than 1 , with their difference is 1. so for $x_{n}$ to be bigger than 2, it is required for $\sqrt{2+\sqrt{3+\sqrt{4+ ... \sqrt{n}}}} \geq 3$ but if square both sides of (1) and substract, we get that $\sqrt{2+\sqrt{3+\sqrt{4+ ... \sqrt{n}}}} \leq 3$.
for (b) , first square both sides, the '1' is gone , square again until the '2' is gone, and we arrive to this equation : $$\sqrt{3+\sqrt{4+...\sqrt{n}}} \leq 2.\sqrt{2+\sqrt{3+...\sqrt{n}}}$$ which is true, because from (1) we know that
$\sqrt{3 +\sqrt{4 ...+\sqrt{n}}} \leq 2$ and $\sqrt{2+\sqrt{3+...\sqrt{n}}} >0$
In fact, if you can prove (b) then (c) is trivial and vice versa.

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