Question: Let C>0 and let a sequence $(x\_i)$ be defined as $x\_1=\sqrt{C}$, and let $x\_{i+1}=\sqrt{C+x\_i}$, prove that it is convergent and find it's limit.

Ok so I have found that the sequence is strictly increasing through induction, but I am having problems showing that it is bounded from above, where then I'd use monotone convergence theorem. I tried using induction but I run into issues no matter what I think of for my bound because it differs if C<1 and if C>1. Any tips?

Answer: 
Consider the upper bound as $1+2\sqrt{C}$ then use induction.

Then by monotone convergence theorem, the limit exist. Then you get a quadratic equation. After solving it and reject the negative root, you can find the limit.
$$x\_n=\underbrace{\sqrt{c+\sqrt{c+...+\sqrt{c}}}}\_\text{n-times}$$
by induction one can show that for all $n$
$$x\_n<1+2\sqrt{c}$$
for $n=1$
$$x\_1=\sqrt{c}<1+2\sqrt{c}$$
assume it is true for $n=k$ then let $n=k+1$. We get (since $c>0$):
$$x\_{k+1}=\sqrt{c+x\_{k}}<\sqrt{c+1+2\sqrt{c}}=\sqrt{(1+\sqrt{c})^2}=1+\sqrt{c}<1+2\sqrt{c}<\infty$$
Now let $\lim\_{n\to\infty}x\_n=L$ then
$$L=\sqrt{c+L}\Rightarrow L^2-L-c=0\Rightarrow L=\frac{1+\sqrt{1+4c}}{2}$$
We ignored the other root as $x\_n>0$ for all $n$.