Asymptotic behavior of recursive sequence Suppose, the real sequence
$x_{n+1}=\frac{1}{2}(x_n+\sqrt{x_n^2+c})$ with c>0 is given.
Find the asymptotic behavior of this sequence.
I have shown, that this sequence goes to infinity as $n\to\infty$ per contradiction. I guess, that it holds
$x_n \approx \frac{1}{2}\sqrt{cn}$
which is motivated by 
$x_{n+1}^2=\frac{1}{2}x_n^2(1+\sqrt{1+\frac{c}{x_n}})+\frac{1}{4}c \approx x_n^2 + \frac{1}{4}c$
where I used the approximation $\sqrt{1+\frac{c}{x_n}} \approx 1$ as $n \to \infty$.
I have problems, to turn this into a formal proof, since it holds for the error term for my approximation
$\frac{1}{2}x_n^2(1-\sqrt{1+\frac{c}{x_n}}) \to \infty$

If we set $y_n=x_n^2$ then
$$y_{n+1}=\frac{y_n}2+\frac c4+\frac{y_n}2\sqrt{1+\frac{c}{y_n}}=y_n+\frac{c}2+O(y_n^{-1}).$$
This proves that $(y_n)$ grows at least linearly, so that $y_n^{-1}=O(n^{-1})$.
Therefore
$$y_n=\frac{cn}2+O(\ln n)$$
and
$$x_n=\sqrt{\frac{cn}2}+O\left(\frac{\ln n}{\sqrt n}\right).$$
