Prove $\sqrt{1+\sqrt{2+\cdots+\sqrt{2006}}} < 2$

[anemone]

Prove that $\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$.

 

[Nono713]

Prove that $\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$.

Spoiler

Note that:
$$\sqrt{c + a} < b ~ ~ \iff ~ ~ c + a < b^2 ~ ~ \iff ~ ~ a < b^2 - c$$
For instance we have:
$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$$
$$\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}<3$$
$$\sqrt{3+\cdots+\sqrt{2006}}<7$$
$$\cdots$$
Unrolling the nested square roots as above, we must therefore show that:
$$\sqrt{2006} < ((((2^2 - 1)^2 - 2)^2 - 3)^2 - \cdots)^2 - 2005$$
Note that the sequence defined by $a_1 = 2$ and $a_{n + 1} = a_n^2 - n$, corresponding to the sequence of RHS above, is increasing. We prove this by induction by showing that $a_n \geq n + 1$. First $a_1 = 2$ so the base case holds, now assume $a_n \geq n + 1$, then $a_n^2 - n \geq (n + 1)^2 - n$, that is, $a_{n + 1} = a_n^2 - n \geq n^2 + n + 1 \geq (n + 1) + 1$ as $n^2 \geq 1$. By induction, $a_n \geq n + 1$ for all $n$. Thus we can now show that $a_{n + 1} = a_n^2 - n \geq a_n$, as that is equivalent to $a_n^2 - a_n - n \geq 0$, and since $m^2 \geq 2m$ for all $\lvert m \rvert > 1$ it follows that $a_n^2 - a_ n - n \geq a_n - n \geq (n + 1) - n \geq 1 \geq 0$ and so $(a_n)_{n = 1}^\infty$ is increasing. Finally, the sequence becomes greater than $\sqrt{2006} \approx 44.788$ at $n = 4$ by simple arithmetic as $a_4 = 46$, which proves the claim. $\blacksquare$​

 

[anemone]

Spoiler

Note that:
$$\sqrt{c + a} < b ~ ~ \iff ~ ~ c + a < b^2 ~ ~ \iff ~ ~ a < b^2 - c$$
For instance we have:
$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$$
$$\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}<3$$
$$\sqrt{3+\cdots+\sqrt{2006}}<7$$
$$\cdots$$
Unrolling the nested square roots as above, we must therefore show that:
$$\sqrt{2006} < ((((2^2 - 1)^2 - 2)^2 - 3)^2 - \cdots)^2 - 2005$$
Note that the sequence defined by $a_1 = 2$ and $a_{n + 1} = a_n^2 - n$, corresponding to the sequence of RHS above, is increasing. We prove this by induction by showing that $a_n \geq n + 1$. First $a_1 = 2$ so the base case holds, now assume $a_n \geq n + 1$, then $a_n^2 - n \geq (n + 1)^2 - n$, that is, $a_{n + 1} = a_n^2 - n \geq n^2 + n + 1 \geq (n + 1) + 1$ as $n^2 \geq 1$. By induction, $a_n \geq n + 1$ for all $n$. Thus we can now show that $a_{n + 1} = a_n^2 - n \geq a_n$, as that is equivalent to $a_n^2 - a_n - n \geq 0$, and since $m^2 \geq 2m$ for all $\lvert m \rvert > 1$ it follows that $a_n^2 - a_ n - n \geq a_n - n \geq (n + 1) - n \geq 1 \geq 0$ and so $(a_n)_{n = 1}^\infty$ is increasing. Finally, the sequence becomes greater than $\sqrt{2006} \approx 44.788$ at $n = 4$ by simple arithmetic as $a_4 = 46$, which proves the claim. $\blacksquare$​

Well done, Bacterius, and thanks for participating!

Solution of other:

Spoiler

Let $X=\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2004+\sqrt{2005+\sqrt{2006}}}}}}$, note that $\sqrt{2005+\sqrt{2006}}<\sqrt{2005+46}<46$

$\therefore X<\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2004+46}}}}$

By the same token

$X<\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2003+46}}}}$
$\vdots$

$X<\sqrt{1+\sqrt{2+\sqrt{3++46}}}=2$ and this completes the proof.