Let $s_{0}$ and $s_{1}$ be arbitrary and $s_{n+1}$=($s_{n} $+ $s_{n-1}$ ) /2 n>=1 show Cauchy

Question

Let $s_{0}$ and $s_{1}$ be arbitrary and $s_{n+1}$=($s_{n} $+ $s_{n-1}$ ) /2 n>=1 show Cauchy show {$s_{n}$} is a Cauchy sequence. Conclude that {$s_{n}$} is convergent.

Answer 1

Observe that $2s_{n+1}=s_n-s_{n-1}$ , thus $2(s_{n+1}-s_n) = -(s_n - s_{n-1})$. If we consider $m\geq n$ we have $$|s_n - s_m| \leq \sum\limits_{i=n}^{m-1} |s_{i+1}-s_i| \leq \frac{1}{2^n} \cdot |s_1-s_0| \sum_{i=0}^{m-n-1} \frac{1}{2^i} \leq \frac{|s_1-s_0|}{2^{n-1}} $$

Therefore (it should be easy to see) $s_n$ in a Cauchy sequence, and since $\mathbb{R}$ is complete, we have that it converges.

Answer 2

Another approach is via difference equations. Your equation in standard form looks as follows: $s_{n+2}-s_{n+1}/2-s_n/2 = 0$. The characteristic equation $\lambda^2-\lambda/2-1/2=0$ has two simple roots 1 and -1/2 so the general solution of the equation is $s_n = c + d (-1/2)^n$ where $c,d \in \mathbb R$. You compute $c,d$ from $s_0,s_1$ to get the concrete solution but you don't need that for the convergence - that's already obvious.

Answer 3

For $n\ge 0,$

$$|s_{n+1}-s_n|=\frac 12|s_n-s_{n-1}|=...$$ $$=\frac{1}{2^{n+1}}|s_1-s_0|$$

for $p>0,$

$$|s_{n+p}-s_n|=\frac{|s_1-s_0|}{2^{n+1}}\Bigl(\frac{1}{2^{p-1}}+\frac{1}{2^{p-2}}+...+\frac 12 +1\Bigr)$$

$$=\frac{|s_1-s_0|}{2^{n+1}}\frac{1-\frac{1}{2^p}}{1-\frac 12}$$

$$=\frac{|s_1-s_0|}{2^n}(1-\frac{1}{2^p})$$ $$<\frac{|s_1-s_0|}{2^n}$$

take now $\epsilon>0$

observe that $$\lim_{n\to+\infty}\frac{|s_1-s_0|}{2^n}=0$$

thus $$(\exists N\in\Bbb N) \;\; : \;\;(\forall n,p\ge N)$$

$$|u_{n+p}-s_n|<\frac{|s_1-s_0|}{2^n}<\epsilon$$ which means that $(s_n)$ is Cauchy.