Due Thursday Sep 9th, 6pm.

In the following, a sequence $(a_n)$ means $(a_n)_{n=0}^\infty$, unless otherwise specified. You can only use properties of real number proved in Tao's book, section 5.4.

1. Let $(a_n)$ be a sequence in $\Q$. Suppose there is a rational $0 < r < 1$, such that $|a_{n+1} - a_n| < r |a_n - a_{n-1}|$, prove that $(a_n)$ is a Cauchy sequence.

2. Let $a_0$ be any positive integer, and $a_{n+1} = 3 + \frac{1}{a_n}$. Prove that the $(a_n)$ is a Cauchy sequence. (Hint: use the previous problem)

3. Let $(a_n)$ be a Cauchy sequence in $\Q$, and let the sequence $(b_n)$ be defined such that $b_n = a_{2n}$. Prove that $(b_n)$ is equivalent to $(a_n)$.

4. If $(a_n)$ and $(b_n)$ are Cauchy sequences, prove that $(a_n b_n)$ is also a Cauchy sequence.

5. If $(a_n)$, $(c_n)$ and $(b_n)$ are Cauchy sequences, and $(a_n) \sim (c_n)$, prove that $(a_n b_n) \sim (c_n b_n)$.

(Problem 4 and 5 together proves Tao proposition 5.3.10, multiplication of real are well-defined).

We first note that, for $n \geq 0$, by repeated using the given inequality, we have $$ |a_{n+1} - a_n| < r^n |a_1 - a_0| $$ We also note that $|a_1-a_0|>(1/r) |a_2-a_1|=0$ by assumption.

Then, for any $n < m$, we have $$ |a_m - a_n| \leq |a_m - a_{m-1}| + |a_{m-1} - a_{m-2}| + \cdots + |a_{n+1} - a_n| \leq |a_1 - a_0| (r^n + \cdots r^{m-1}) \leq |a_1 - a_0| \frac {r^n}{1-r}. $$

For any $\epsilon>0$, we may take integer $N$ large enough, such that $$ |a_1 - a_0| \frac {r^N}{1-r} < \epsilon $$ Thus, for all $n,m > N$, we have $$ |a_m - a_n| \leq |a_1 - a_0| \frac {r^n}{1-r} \leq |a_1 - a_0| \frac {r^N}{1-r} \leq \epsilon. $$

For integer $n \geq 1$, we have $$ a_{n+1} - a_n = (3 + \frac{1}{a_n}) - (3 + \frac{1}{a_{n-1}}) = \frac{a_{n-1} - a_n}{a_n a_{n-1}} $$