prove that $x_{n+1}=c+\sqrt{x_n}$ How can I prove that $x_{n+1}=c+\sqrt{x_n}$, $x_1=a>0$ and $c>0$ converges?
I know that the limit (if it exists) is $L={{2c+1+\sqrt{4c+1}}\over 2}$.
I have already prove that if $x_1<L$ then $x_n<L$ so its bounded from above but how can I prove that if $x_1<L$ then the sequence is increasing? 
I would really appreciate any hints or ideas.

So you have $x_{n+1}-x_n=c+\sqrt{x_n}-x_n$. Now consider the function:
$$f(x)=c+\sqrt{x}-x,x\in R.$$
It follows
$$f'(x)=\frac{1}{2\sqrt{x}}-1.$$
You can see that when $x< \frac{1}{4}$, it has
$$f'(x)>0.$$
This means
$$f(x)>f(\frac{1}{4})=1+\sqrt{\frac{1}{4}}-\frac{1}{4}=\frac{5}{4}>0.$$
That's
$$x_{n+1}-x_n>0,n=1,2,\cdots, \text{ if } x< \frac{1}{4}.$$
Conversly, Let $f(x)>0$ and $f'(x)>0$, you can easily get the number you want $L=\frac{2c+1+\sqrt{4c+1}}{2}$.
