Showing $s_n = \left(\frac{1}{2}\right)(s_{n-1} + s_{n-2})$ is Cauchy.

Question

Let $(s_n)$ be a sequence defined as $s_1 = 1, s_2 = 2$ , and $s_n = \left(\frac{1}{2}\right)(s_{n-1} + s_{n-2})$. Prove that $(s_n)$ is Cauchy.

I can see how it is convergent and Cauchy but not sure how to put it into a formal proof.

Answer 1

We have

\begin{align} s_n - s_{n-1} &= \dfrac{s_{n-1}+s_{n-2}}{2} - s_{n-1} \\ &= -\dfrac{1}{2}(s_{n-1}-s_{n-2}) \\ &= \cdots \\ &=(-\dfrac{1}{2})^{n-2}(s_2-s_1) \\ \end{align}

Let $M=\dfrac{4}{3}|(s_2-s_1)|$, for any $\epsilon$, set $\space N=\log_2(\dfrac{M}{\epsilon})+1$, then for any $n>m, \space m>N,\space m,n\in \mathbb{N}$, there is

\begin{align} |s_n - s_m| &= |(s_{n}-s_{n-1})+\cdots+(s_{m+1}-s_{m})| \\ &=|(-\dfrac{1}{2})^{n-2}+\cdots+(-\dfrac{1}{2})^{m-1}||(s_2-s_1)| \\ &= \left|\dfrac{(-\dfrac{1}{2})^{m-1}(1-(-\dfrac{1}{2})^{n-m})}{1+\dfrac{1}{2}}\right||(s_2-s_1)| \\ &=\dfrac{2}{3}\dfrac{1}{2^{m-1}}\left|(1-(-\dfrac{1}{2})^{n-m})\right||(s_2-s_1)| \\ &<\dfrac{M}{2^{m-1}} \\ &<\epsilon \end{align}

So $s_n$ is Cauchy sequence.

Answer 2

$$s_n - s_{n-1} = \dfrac{s_{n-1}+s_{n-2}}{2} - s_{n-1} = -\dfrac{1}{2}\left(s_{n-1}-s_{n-2}\right) = \left(-\dfrac{1}{2}\right)^2\left(s_{n-2}-s_{n-3}\right) = \cdots = \left(-\dfrac{1}{2}\right)^{n-2}(s_2-s_1) = \left(-\dfrac{1}{2}\right)^{n-2} \Rightarrow s_n = (s_n-s_{n-1})+(s_{n-1}-s_{n-2})+\cdots + (s_2-s_1)+s_1 = \left(-\dfrac{1}{2}\right)^{n-2} + \left(-\dfrac{1}{2}\right)^{n-3}+\cdots + (2-1)+1$$. You can now simplify this expression and get a closed form formula for $s_n$, and you can proceed to show it is a Cauchy sequence.