Proof a Cauchy sequence.

Question

Given is a sequence $(a_{n})_{n \in N}$ and is defined as $a_{n+1} - a_{n} = \frac{1}{2}(a_{n-1} -a_{n})$. I have to proof that this is a Chauchy sequence, but i don't know how to start. I know that a Cauchy row is defined as for a $ \forall \epsilon > 0: \exists n_{0} \in N$: $ \forall n,m \geq n_{0}$ that $ |a_{n} - a_{m}| $ < $ \epsilon $. But how do I proof for an random Epsilon that it is smaller? Thanks!

Answer 1

We get:

$$a_{n+1} - a_n = \frac {-1}2 (a_n-a_{n-1}) = \frac 14(a_{n-1}-a_{n-2}) = \frac {-1}{2^3}(a_{n-2}-a_{n-3})= \cdots = \frac{(-1)^n}{2^n}(a_1-a_0)$$

Hence for $n>N$ we get that:

$$|a_n - a_N| \le |a_n-a_{n-1}|+|a_{n-1}-a_{n-2}|+ \cdots +|a_{N+1}-a_N| = (a_1-a_0)\sum_{i=N-1}^{n-1} \left(\frac{1}{2}\right)^i$$ $$ = |a_1-a_0| \left(\frac{1}{2}\right)^{N-1}\frac{1-\left(\frac{1}{2}\right)^{n-N}}{1-\frac{1}{2}} = |a_1-a_0|\left(\frac 12 \right)^{N-2} \left(1-\left(\frac{1}{2}\right)^{n-N}\right)< |a_1-a_0|\left(\frac 12 \right)^{N-2}$$

Now obviously for any $\epsilon>0$ we can choose suitable $N$ s.t. $(a_1-a_0) \left(\frac{1}{2}\right)^{N-3} < \epsilon$. Then:

$$|a_n-a_m| \le |a_n-a_N| + |a_m-a_N| <|a_1-a_0|\left(\frac 12 \right)^{N-2} + |a_1-a_0|\left(\frac 12 \right)^{N-2} = |a_1-a_0|\left(\frac 12 \right)^{N-3} < \epsilon$$

Therfore $(a_n)$ is a Cauchy sequence.