How do I prove this sequence is always smaller than this following limit

Question

How do I prove this sequence, which I found is monotonic and rising

$ a_{1} = 1, a_{n+1} = \sqrt{c+da_{n}}$
while $d,c >1 $

is always smaller than the suspicious limit I found to it $L = \frac{d+ \sqrt{d^2+4c}}{2}$

tried induction but no luck.

Answer 1

Once you already know that $a_n$ is monotonic increasing then $a_{n-1}

$$a_n^2=c+da_{n-1}

Answer 2

hint

Let $f(x)=\sqrt{c+dx}$ for $x\geq 1$.

$f'(x)=\frac{d}{2f(x)}>0$ $f$ is strictly increasing at $[1,+\infty)$.

$L=\frac{d+\sqrt{d^2+4c}}{2}>1$ is a fixed point of $f$.

by induction , it is easy to prove that $\forall n>0 \; a_n

• $a_1=1

• $a_n