Prove that $a_{n}$ converges for $a_{1} \ge 2$ and $a_{n+1} = 1+ \sqrt{a_{n}-1}$ and find its limit Prove that $a_{n}$ converges for $a_{1} \ge 2$ and $a_{n+1} = 1+ \sqrt{a_{n}-1}$ and find it's limit. First of all, what i did is to find the first four terms of the sequence.
$a_{2} = 1+ \sqrt{a_{1}-1} \ge 2$
$a_{3} = 1+ \sqrt{a_{2}-1} = 1+ \sqrt{\sqrt{a_{1}-1}} \ge 2$
$a_{4} = 1+ \sqrt{\sqrt{\sqrt{a_{1}-1}}} \ge 2$
Then I assume $a_{n} \ge 2 $ Witch means that $a_{n}$ it's lower bounded. I need to prove that it is a decreasing sequence and then $a_{n}$ would converge.
$$\frac{a_{n+1}}{a_{n}} = \frac{1+\sqrt{a_{n}-1}}{a_{n}} \le1$$ but i don't know how to prove that.  The limit, I don't know how to find it.
Thanks in advance.
 A: Hint
Set $b_n=a_n-1$ ($b_1\ge 1$), then we have:
$$b_{n+1}=\sqrt{b_n}.$$
Also see that, $$b_2=b_1^{1/2}$$ $$b_3=b_2^{1/2}=b_1^{1/4}$$ $$b_4=b_3^{1/2}=b_1^{1/8}$$ $$...$$
can you finish?
A: 
Then I assume $a_n\ge 2$

Note that you would have to prove that as well. Just because the first four terms are $\ge 2$, doesn't mean that you can conclude that for all $n$. However, induction should easily help you out there.
If you assume $a_n \ge 2$, then you get
$$a_{n+1} = 1 + \sqrt{\color{blue}{a_n} - 1} \ge 1 + \sqrt{\color{blue}{2} - 1} = 2.$$
Now, since $a_1 \ge 2$, induction tells you that $a_n \ge 2$ for all $n$.

Now, showing that $a_n$ is decreasing is easy because we have
$$a_n - 1 \ge 1$$
and thus,
$$a_n - 1 \ge \sqrt{a_n - 1}.$$
Rearranging the above gives us that
$$a_n \ge 1 + \sqrt{a_n - 1} \ge a_{n+1}.$$
This finishes the proof the way you intended it.

Note that the other answer gives a simpler alternative by considering $b_n = a_n - 1$.

Once you know that the sequence converges, you can find the limit as following:
$$a_{n+1} = 1 + \sqrt{a_n - 1} \implies \lim_{n\to\infty}a_{n+1} = \lim_{n\to\infty}1 + \sqrt{a_n - 1}.$$
If $\displaystyle\lim_{n\to\infty}a_n = A$, then $\displaystyle\lim_{n\to\infty}a_{n+1} = A$ as well.
Moreover, $\sqrt{.}$ is continuous and thus, you may "take the limit inside". Thus, the above equation tells you
$$A = 1 + \sqrt{A - 1}$$
or $$A - 1 = \sqrt{A - 1}.$$
Squaring the above and solving it gives two solutions: $A = 1, 2$.
However, note that $a_n \ge 2$ for all $n$ and thus, $A \ge 2$ as well. With this, you can conclude that
$$\lim_{n\to\infty}a_n = \boxed{2}.$$
A: Let prove it is decreasing
$$a_{n+1}-a_n=$$
$$1+\sqrt{a_n-1}-a_n=$$
$$\sqrt{a_n-1}(1-\sqrt{a_n-1})$$
but
$$a_n\ge 2\implies \sqrt{a_n-1}\ge 1$$
$$\implies 1-\sqrt{a_n-1}\le 0$$
$$\implies a_{n+1}-a_n\le 0$$
thus $ (a_n) $ is decreasing and bounded below, so it converges to $ L $ satisfying
$$L=1+\sqrt{L-1}\implies L\in\{1,2\}$$
$$\implies L=2$$
