$a_0=\sqrt k$
$a_{n+1}:=\sqrt{a_n+k}$
I am trying to show the convergence by monotone convergence theorem and I can show its monotonicity by observing the square of each term but I am stuck at proving its boundedness. I feel it should be bounded as it increases slower and slower.
What is $a_1$? Assume $x_0$ is positive root of $x^2-x-k=0$ (Assume $k\gt 0$)
If $a_1 \gt x_0$, we could prove $x_0\lt a_{n+1}\lt a_n$
since $a_2 = \sqrt{a_1+k}$, so $a_1^2-a_2^2=a_1^2-a_1-k\gt 0$, so $a_2\lt g_1$ and $(a_1+\sqrt{a_1+k})(a_2^2-a_2-k)=(a_1+\sqrt{a_1+k})(a_1-\sqrt{a_1+k})=a_1^2-a_1-k\gt 0$, so $a_2^2-a_2-k\gt 0$, so $a_2\gt x_0$.
So our assumption holds for n=2
Similarly, assuming $x_0\lt a_{m+1} \lt a_m$
We have $a_{m+1}^2-a_{m+2}^2=a_{m+1}^2-a_{m+1}-k \ gt 0$, so $a_{m+2} \lt a_{m+1}$
and $(a_{m+1}+\sqrt{a_{m+1}+k})(a_{m+2}^2-a_{m+2}-k)=a_{m+1}^2-a_{m+1}-k \ gt 0$,
so $a_{m+2}^2-a_{m+2}-k \ gt 0$, so $a_{m+2} \gt x_0$
So $x_0\lt a_{n+1}\lt a_n$ is true for any n.
Similarly for $a_1 \lt x_0$
