1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $1+\sqrt 2 x_n<1+2\sqrt 2<4$

$\endgroup$
1
$\begingroup$

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...

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.