Showing that this particular sequence is Cauchy using the definition

Question

This is the following information that I've been given;

$$a_{1}=1,\, a_{2}=4,\, a_{n+2}=\frac{2a_{n+1}+a_{n}}{3}$$

I need to show that $|a_{m}-a_{n}|<\epsilon \,,\forall n\geq n_{0}, m>n$

I deconstructed the general sequence to the following $$|a_{m}-a_{m-1}|+|a_{m-1}-a_{m-2}|+...+|a_{n+2}-a_{n+1}|+|a_{n+1}-a_{n}|$$ The above method was what our tutor showed us when basically telling what we would need to know to solve this, although I'm having difficulty getting from that back to the inequality to show that this sequence is or isn't Cauchy.

Answer 1

Observe that your recursive relation implies $$ |a_{n+2}-a_{n+1}|= \frac 1{3}|a_{n+1}-a_{n}| $$ So the differences are decreasing geometrically with ratio $1/3$. That is, for $m\le n$ $$ S_{mn}=|a_m-a_{m-1}|+|a_{m-1}-a_{m-2}|+\dots |a_{n+1}-a_{n}|=(1+1/3+1/9+\dots +1/3^{n-m})|a_{n+1}-a_{n}|. $$ Now, given $\epsilon$ choose $n$ large such that $|a_{n+1}-a_{n}|<2\epsilon/3$ (this is possible because the sequence of the differences is geometric and decreasing to zero). Then we conclude $$ S_{mn}\le 2(1+1/3+1/9+\dots +1/3^{n-m})\epsilon/3\le \epsilon. $$ for all $m>n$ since $$ 1+1/3+1/9+\dots +1/3^{n-m}\le \sum_{i=0}^\infty 3^{-i}=3/2. $$ Finally, just observe that $$ |a_m-a_n|\le S_{mn} $$ by the triangle inequality.

Answer 2

More generally, if $a_{n+2} = ca_{n+1}+(1-c)a_n$ where $0 < c < 1$ then $a_{n+2}-a_{n+1} =ca_{n+1}+(1-c)a_n-a_{n+1} =(c-1)a_{n+1}+(1-c)a_n =(c-1)(a_{n+1}-a_n) $ so $|a_{n+1}-a_n|$ decreases in absolute value and alternates in sign.

$a_n$ then converges in the same way as shown in the other answer.

Answer 3

$$3(a_{n+2}-a_{n+1})=a_n-a_{n+1}.$$ Let $b_n=a_{n+1}-a_n.$

Thus, $$b_{n+1}=-\frac{1}{3}b_n,$$ which says $$a_{n+1}-a_n=b_n=b_1\left(-\frac{1}{3}\right)^{n-1}=3\left(-\frac{1}{3}\right)^{n-1}=-\frac{1}{(-3)^{n-2}}.$$ Now, $$|a_m-a_n|=\left|\sum_{k=m-1}^n(a_{k+1}-a_k)\right|=\left|\sum_{k=m-1}^n\frac{1}{(-3)^{k-2}}\right|<\frac{1}{3^{n-2}}.$$ Can you end it now?