For $c > 1$, define sequence $\{a_n\}$ with $a_1 = \sqrt{c(c-1)}$ and $a_{n+1} = \sqrt{c(c-1) + a_n}$, $n > 1$. Prove that $\{a_n\}$ is a monotonically increasing sequence and it's bounded by $c$.
My attempt:
Obviously $a_1 < a_2$, by induction, suppose $a_n > a_{n-1}$, i.e. $a_n - a_{n-1} > 0$. We have $a_{n+1}^2 - a_n^2 = c(c-1) + a_n - c(c-1) - a_{n-1} = a_n - a_{n-1} > 0$. Hence, $\{a_n\}$ is a monotonically increasing sequence. Otherwise, $a_1 = \sqrt{c(c-1)} \leq c$ because $c^2 - a_1^2 = c^2 - c(c-1) = c \geq 0$, by induction, suppose $a_n \leq c$, we prove that $a_{n+1} \leq c$. $a_{n+1} = \sqrt{c(c-1) + a_n} < \sqrt{c(c-1) + c}$, but $c^2 - \left(\sqrt{c(c-1) + c}\right)^2 \geq 0$, then $a_{n+1} \leq c$.
Is this solution true? Thank all!
