Investigate the convergence or divergence of the sequence $a_1 = \sqrt{c}, a_2 = \sqrt{c + a_1}, a_{n+1} = \sqrt{c + a_n}$ [duplicate]

Claim: $a_n$ is increasing and bounded above by $M = \dfrac{1+\sqrt{1+4c}}{2}$. Also $\displaystyle \lim_{n\to \infty} a_n = M$

Proof: By induction on $n \ge 1$. $a_2 = \sqrt{c+\sqrt{c}}> \sqrt{c} = a_1$. Assume $a_n > a_{n-1}$, we show $a_{n+1} > a_n$. But $a_{n+1} = \sqrt{c+a_n}> \sqrt{c+a_{n-1}}= a_n$ by the inductive step. So by induction $a_{n+1} > a_n, \forall n \ge 1$. We show that $a_n < M, \forall n \ge 1$ by induction also. For $n = 1, a_1 = \sqrt{c} < M \iff \sqrt{c} < \dfrac{1+\sqrt{1+4c}}{2}\iff 2\sqrt{c} - 1 < \sqrt{1+4c}$. If $c < \dfrac{1}{4}$, we're done since $LHS < 0 < RHS$. Otherwise, $c > \dfrac{1}{4}\implies 2\sqrt{c} - 1 > 0$, and squaring both sides: $4c - 4\sqrt{c} + 1 < 1+ 4c \iff -4\sqrt{c} < 0$ is clearly true. Thus $a_1 < M$. Assume $a_n < M$, you have $a_{n+1} = \sqrt{c+a_n} < M \iff a_n < M^2 - c = \dfrac{1+2\sqrt{1+4c}+1+4c}{4} -c = \dfrac{1+\sqrt{1+4c}}{2} = M$, which is true by inductive step. So $a_n < M, \forall n \ge 1$. Thus the claim is verified. This follows that the limit exists and call it $M$. Then $M = \sqrt{c+M}\implies M^2 = c+M\implies M^2-M -c = 0\implies M = \dfrac{1 \pm \sqrt{1+4c}}{2}$. Since $a_n > 0,\forall n \ge 1 \implies M \ge 0\implies M = \dfrac{1+\sqrt{1+4c}}{2}$ .

First, you need to study the fixed points of the function $$ f(x) = \sqrt{c+x} $$ as only these can be the limit of the sequence $f(0), f(f(0)), f(f(f(0))), \dots$.

Solvind $y=\sqrt{c+y}$, we find $$ y_{\pm}=\frac{1\pm \sqrt{1+4c}}{2} $$

The sequence starts at $x=0$ and is nonnegative, so the only possible candidate for the limit is $$ y=\frac{1 + \sqrt{1+4c}}{2} $$

Now, we need to prove that if $0<x<y$ then $x<f(x)<y$. $$ x<f(x) = \sqrt{c+x}<\sqrt{c+y}=f(y)=y $$ Thus, the sequence is bounded from above by $y$ and is increasing. As $y$ is the only possible limit, it converges to $y$.