Let $(x_n)^{\infty}_{n=1}$ be a sequence in $\mathbb{R}$ such that there is θ ∈ [0, 1) with $|x_{n+1} − x_n| ≤ θ^n$ for n ∈ N.
Show that $x_n$ is a Cauchy sequence and therefore converges.
Attempt:
claim: $x_n$ is cauchy
Then there exists $|x_m-x_n|$ < E , where $E>0$
However, since θ^n can be made equal to 0, when $θ \in[0,1)$ we cannot construct an E>0 therefore the sequence is not cauchy.
On
Hint. Let's bound the amount $|x_n-x_m|$. Without loss of generality, assume $m>n$. Then $m=n+k$ for some $k\geq1$. Now, see the following $$\begin{align} |x_n-x_m| &= |x_n-x_{n+k}| \\ &\leq |x_n-x_{n+1}|+|x_{n+1}-x_{n+2}|+\cdots+|x_{n+k-1}-x_{n+k}| \\ &\leq \theta^n+\theta^{n+1}+\cdots+\theta^{n+k-1} \\ &< \sum_{j=n}^\infty \theta^j = \frac{\theta^n}{1-\theta} \end{align}$$
First, when we say that there is an number $\theta \in [0,1)$ such that blah blah blah, we mean that $ \theta $ is a fixed number like any other number in [0,1), as, for example, 1/2. Therefore, you can't take $\theta$ in any way you want to make your proof correct, but you need suppose that the fixed $\theta$ just lies in [0,1).
Second, your main objetive is to prove that $\{x_k\}$ is Cauchy provided that there exists a fixed $\mathbf{\theta}$ $\in (0,1]$ satisfying $$|x_{n+1} − x_n| ≤ {\theta}^n$$ for every natural number $n$. Aiming that, you need to prove that for every $\epsilon$ greater than zero (not depending on whether it's too big or too close to zero) there exists a natural number $n_{\epsilon}$, i.e., depending just on $\epsilon$, such that for every pair of natural numbers $m,n >n_{\epsilon}$ $$ |x_n - x_m|< \epsilon.$$ Because of that, you can't suppose that there exists $|x_m−x_n| < \epsilon$ at the start of your proof, but you need start with the hypoteses you have, i.e,
$$|x_{n+1} − x_n| ≤ {\theta}^n$$ for every natural number $n$. Saying all of this, your proof will look like follows.
Proposition:Let $\{x_k\}$ be a sequence of real numbers such that there exists a fixed $\mathbf{\theta}$ $\in (0,1]$ satisfying $$|x_{n+1} − x_n| ≤ {\theta}^n$$ for every natural number $n$, then $\{x_n\}$ is Cauchy.
Proof: Suppose that there exists a fixed $\mathbf{\theta}$ $\in (0,1]$ satisfying $$|x_{n+1} − x_n| ≤ {\theta}^n$$ for every natural number $n$. Then, for every $\epsilon>0$, we can chose $n_0\in\mathbb{N}$ such that for every $n>n_0$ follows that $\frac{\theta^n}{1-\theta} \leq \epsilon$, because $$\lim_{n\rightarrow \infty} \frac{\theta^n}{1-\theta} = 0. $$ Following that, observe that for every pair of positive natural numbers $n, m> n_0$, setting first the case $m>n$, we have $$\begin{align} |x_n-x_m| &= |x_n-x_{n+k}| \\ &\leq |x_n-x_{n+1}|+|x_{n+1}-x_{n+2}|+\cdots+|x_{n+k-1}-x_{n+k}| \\ &\leq \theta^n+\theta^{n+1}+\cdots+\theta^{n+k-1} \\ &< \sum_{j=n}^\infty \theta^j = \frac{\theta^n}{1-\theta} \end{align}.$$ Then $|x_n-x_m|\leq \epsilon$ for each pair of numbers $n>m>n_0$. For the case $n>m>n_0,$ note that it's just exchange places of $m,n$ in the previous inequations and we will have also $|x_n-x_m|<\epsilon.$ Hence, the sequence is Cauchy.