Consider the sequence $\{a_n\}$ defined as $a_1=0$ and for $n>1$, $\displaystyle a_{n+1}=\frac{(a_n+3)^3}{100}$. Check whether the sequence is convergent or not.
My initial guess is that the sequence is converegent. I want to check that sequence is monotone and bounded. We have, \begin{align*} a_{n+1}-a_n=\frac 1{100}(a_n^3+9a_n^2-73a^n+27) \end{align*} From this I'm unable to decide the sign of the right hand side so that we can say about increasing or decreasing.
How to proceed ? Any other prominant way to check the convergence ?
You can show (by induction) that the sequence is increasing and bounded.
First boundedness: $$0\leq a_n <1 \Rightarrow 0\leq a_{n+1} =\frac{(a_n+3)^3}{100}< \frac{4^3}{100}<1$$
Increasing: $$a_{n+2}-a_{n+1} = \frac 1{100}\left((a_{n+1}+3)^3-(a_{n}+3)^3\right)$$$$=\frac{a_{n+1}-a_n}{100}((a_{n+1}+3)^2 + (a_{n+1}+3)(a_{n}+3) + (a_{n}+3)^2)\geq 0$$
Since $a_2-a_1 \geq 0$, monotonicity follows by induction. So, $a_n$ is convergent.