18. Sequences

The first \(4\) terms are: \[\begin{aligned} a_1&=4 \\ a_2&=\dfrac{1}{8}4^2+\dfrac{3}{2}=\dfrac{7}{2}=3.5 \\ a_3&=\dfrac{1}{8}\left(\dfrac{7}{2}\right)^2+\dfrac{3}{2} =\dfrac{97}{32}\approx3.03 \\ a_4&=\dfrac{1}{8}\left(\dfrac{97}{32}\right)^2+\dfrac{3}{2} =\dfrac{21\,697}{8192}\approx2.65 \end{aligned}\] Before showing the limit exists, we want to get some idea what the limit might be. So we assume the limit exists and try to find it.

2.   Assume the sequence has a limit \(L\). Then \(\lim\limits_{n\to\infty}a_n=L\) and also \(\lim\limits_{n\to\infty}a_{n+1}=L\) since \(a_{n+1}\) is the same sequence as \(a_n\) but starting from a different number. We apply \(\lim\limits_{n\to\infty}\) to both sides of the recursion relation and use the limit laws: \[\begin{aligned} a_{n+1}&=\dfrac{1}{8}(a_n)^2+\dfrac{3}{2} \\ \lim_{n\to\infty}a_{n+1} &=\dfrac{1}{8}\left(\lim_{n\to\infty}a_n\right)^2+\dfrac{3}{2} \\ L&=\dfrac{1}{8}L^2+\dfrac{3}{2} \end{aligned}\] We can solve this for \(L\): \[\begin{aligned} 8L=L^2+12 \quad \Longto \quad &L^2-8L+12=0 \quad \Longto \\ (L-2)(L-6)=0 \quad &\Longto \quad L=2 \quad \text{or} \quad 6 \end{aligned}\] This says that if a limit exists, it must be either \(2\) or \(6\). Looking at the first five terms, we expect the limit is \(2\).

We now turn to showing the limit actually exists.

1.   To show the limit exists, we use the Bounded, Monotonic Sequence Theorem. Looking at the first five terms, we expect the sequence is decreasing and bounded below by \(2\). To prove each of these we will use mathematical induction:

We have \(a_1=4\gt2\). Suppose for some \(k\) we have \(a_k\gt2\). Then \((a_k)^2\gt4\) and \(a_{k+1}=\dfrac{1}{8}(a_n)^2+\dfrac{3}{2} \gt\dfrac{1}{8}4+\dfrac{3}{2}=2\). By mathematical induction, \(a_n\gt2\) for all \(n\).

Notice the first four terms are decreasing. Suppose for some \(k\) we have \(a_k>a_{k+1}\). Then \((a_k)^2>(a_{k+1})^2\) and \(\dfrac{1}{8}(a_k)^2+\dfrac{3}{2} >\dfrac{1}{8}(a_{k+1})^2+\dfrac{3}{2}\) which says \(a_{k+1}>a_{k+2}\). By mathematical induction, the sequence is decreasing.

Therefore: The sequence has a limit and the limit must be \(2\).

The first \(3\) terms are: \[\begin{aligned} a_1&=8 \\ a_2&=\dfrac{1}{8}8^2+\dfrac{3}{2}=\dfrac{19}{2}=9.5 \\ a_3&=\dfrac{1}{8}\left(\dfrac{19}{2}\right)^2+\dfrac{3}{2} =\dfrac{409}{32}\approx12.78 \end{aligned}\] Before showing the limit exists, we want to get some idea what the limit might be. So we assume the limit exists and try to find it.

2.   Since the recurrance relation is the same as in the previous example, the process of applying \(\lim\limits_{n\to\infty}\) is also the same and we conclude if a limit exists, it must be either \(2\) or \(6\). However, looking at the first \(3\) terms, we see that the terms start at 8 and seem to be increasing. If we can prove it continues to increase, there will not be a limit.

Notice the first three terms are increasing. Suppose for some \(k\) we have \(a_k< a_{k+1}\). Then \((a_k)^2<(a_{k+1})^2\) and \(\dfrac{1}{8}(a_k)^2+\dfrac{3}{2} <\dfrac{1}{8}(a_{k+1})^2+\dfrac{3}{2}\) which says \(a_{k+1}< a_{k+2}\). By mathematical induction, the sequence is increasing.

We now ask if it is bounded above. If there was an upper bound, the Bounded, Monotonic Sequence Theorem would say there is a limit. But since the sequence is increasing from \(8\), that limit would be bigger than \(8\) which is impossible since the limit has to be either \(2\) or \(6\). So there is no upper bound. By the Bounded, Monotonic Sequence Theorem, we know the sequence diverges to \(\infty\).