Investigate the convergence or divergence of the sequence a1=c,a2=c+a1,an+1=c+an,c>0. I proved it is increasing. But...

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

First, you need to study the fixed points of the function
${\displaystyle f\left(x\right)=\sqrt{c+x}}$
as only these can be the limit of the sequence
${\displaystyle f\left(0\right),f\left(f\left(0\right)\right),f\left(f\left(f\left(0\right)\right)\right),\dots }$
Solving ${\displaystyle y=\sqrt{c+y}}$, we find
${\displaystyle y_{\pm }=\frac{1\pm \sqrt{1+4c}}{2}}$
The sequence starts at ${\displaystyle x=0}$ and is nonnegative, so the only possible candidate for the limit is
${\displaystyle y=\frac{1+\sqrt{1+4c}}{2}}$
Now, we need to prove that if ${\displaystyle 0<x<y\text{then}x<f\left(x\right)<y}$.
${\displaystyle x<f\left(x\right)=\sqrt{c+x}<\sqrt{c+y}=f\left(y\right)=y}$
Thus, the sequence is bounded from above by y and is increasing. As y is the only possible limit, it converges to y.