The given sequence is:
{ $a_n$} $= \sqrt{2+\sqrt{2+...+\sqrt{2}}}$, where n is the number of square roots, or in other words, $\{a_n\} = \sqrt{2+\{a_{n-1}\}}$
In the first part, you solve to determine that { $a_{n+1}$} > { $a_n$}
In the next part of the problem, it says to show by inducion that { $a_n$} is bounded above by 2 and then to find the $lim_{n\to\infty}${ $a_n$}. But then I looked for sample problems in my workbook and found this exact question but it had said that it should be bounded by 3, not 2.

Although it explicitly states that it should be 3, I can only get 2, I was wondering if I made an error or was it maybe a misprint in my book?

{ $a_n$} < 2 for all n = 1, 2, 3...
Assume that it is true for n = k, that is, { $a_k$} < 2. Then,
$\{a_{k+1}\} < 2$
$\sqrt{2+\{a_k\}}< 2$
However, $\{a_k\} < 2$
so $\sqrt{2+\{a_k\}} < \sqrt{2 + 2} < 2$
Therefore $\{a_k\}$ is always less than two.

Then since it has been proved that this is a bounded monotonic sequence, it will converge to some value $\lim_{n\to\infty}\{a_n\} = L$
This is how I solved:
$\lim_{n\to\infty}\{a_{n+1}\} = L$

$\lim_{n\to\infty}\sqrt{2+\{a_k\}} = L$

$\frac{1}{2} \lim_{n\to\infty} (2+\{a_k\}) = L$

$\frac{1}{2}(2 + L) = L$
and then L = 2, so the sequence converges to 2.

2. $Let y = sqrt(y+2)$ as n tends to infinity

$y^2 = y+2$
$y^2-y-2=0$
solving for y,
we get y=2 or y=-1,(which is impossible.)

So the limit of the sequence should be 2.