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.

Note that we have
$$x_{n+1}-x_n=\sqrt{x_n}-\sqrt{x_{n-1}}=\frac{x_n-x_{n-1}}{\sqrt{x_n}+\sqrt{x_{n-1}}}$$
Hence, if $c+\sqrt{x_1}>x_1$, then we see from $(1)$ by inductive reasoning that the sequence is always increasing.
Similarly, if $c+\sqrt{x_1}<x_1$, then the sequence is decreasing. 
