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: 
$x\_1 \leq 1+2\sqrt{C}$, and
$x\_i \leq 1+2\sqrt{C} \to x\_{i+1}\leq \sqrt{C+2\sqrt{C}+1}=\sqrt{C}+1\leq 1+2\sqrt{C}$.
Let $f(x) = \sqrt{C+x}$, and note that $f$ is increasing on $[-C,\infty)$.

Let $\phi(x) = f(x)-x$, note that $\phi(0) >0$, and $\lim\_{x \to \infty} \phi(x) = -\infty$.

Let $x^\* $ be the smallest $x$ such that $\phi(x) = 0$. We have $x^\* = f(x^\*)$, squaring gives $(x^\*)^2-(x^\*)-C = 0$, from which we get $x^\*= {1 \over 2} (1 + \sqrt{1+4C})$.

Then if $0 \le x < x^\*$, $\phi(x)>0$ and $f(x) < f(x^\*)$, so $x < f(x) < f(x^\*)=x^\*$.

Hence if $x\_0 < x^\*$, then induction shows that $x\_n < x\_{n+1} < x^\*$. Consequently, $x\_n$ has a limit $\hat{x}$, and since
$x\_{n+1} = f(x\_n)$, we see that $\hat{x} = f(\hat{x})$, and since $x\_n < x^\*$, we see that $\hat{x} \le x^\*$ and so $ \hat{x} = x^\*$ (since it is the smallest such $x$).