Show that the function converges to the following limit.

Question

Consider the sequence $a_1=-\frac{3}{2}$ and $3a_{n+1}=2+a_n^3$.I have to show that $(a_n)\to 1$ as $n\to \infty$.I know I have to show monotonicity and boundedness but I cannot find a proper bound for the sequence.Can someone help me.I also want motivation behind the solution.

Answer 1

The first few values of this sequence are $$\begin{align} &-1.5,\\&-0.45833333,\\&0.63457272,\\&0.75184379,\\&0.80833135, \\&\dots \end{align}$$

Based on this we make the guess $|a_n|\leq 1, n \geq 2$ (any finite initial part of the sequence can be discarded, they do not contribute to the convergence, and since there are a finite number of elements in such a part, it is bounded as well). Now we can prove by induction that this guess is correct. Indeed $|a_2| = 0.45\dots \leq 1.$ Now if $|a_n|\leq 1,$ then, using the recursive formula we find \begin{align*} |a_{n+1}| & = \frac{1}{3} | 2+a_n^3|\\ &\leq \frac{1}{3}(2+ |a_n|^3)\\ &\leq 1. \end{align*} Hence the sequence is bounded by $\max\{|a_1|,1\}.$

Similarly we can show that $a_n >0$ for $n\geq 3.$ Hence we have for $n\geq 3,$ that $0 < a_n \leq 1.$

To show monotonicity, we will check that $a_{n+1} - a_n >0.$ Using the recursive formula this is the same as showing that $$ \frac{1}{3}(2+a_n^3) - a_n >0. $$

This follows easy by considering the function $x\mapsto \frac{1}{3}\left(2+x^3\right)-x$ on the interval $[0,1].$

Hence we have that the sequence $\left(a_n\right)_{n}$ is bounded and monotonically increasing, hence it is convergent.

To find the limit, let $L = \lim\limits_{n\to\infty} a_n$ and take the limits in the recursive relation to get $$ 3 L = 2 + L^3. $$ Solving for $L$ yields $L \in \{-2,1\}$ and by the above analysis, $-2$ is not an option, so $L= 1.$