Prove that a sequence $\left\{a_{n}\right\}$ that satisfies $\left|a_{n+1}-a_{n}\right|
Show that any sequence $\left\{a_{n}\right\}$ that has the property $\left|a_{n+1}-a_{n}\right|<b^{n}$ for $b<1$ is a Cauchy sequence.
(proof)
Let $ b\in (-1,1) $.
Given any two points of the sequence $ a_n $ and $ a_{n+k} $ the distance between them is 
\begin{align*}
    |a_{n+k}-a_n| &\leq  |a_{n+1}-a_n|+|a_{n+2}-a_{n+1}|+...+|a_{n+k}-a_{n+k-1}| \\
    &< b^n + b^{n+1} + b^{n+2} + \cdots + b^{n+k-1}\\
    &= \sum_{i=0}^{k-1}b^{n+i}\\
    &\leq \sum_{i=0}^{k-1}|b^{n+i}|\\
    &= |b^n|\sum_{i=0}^{k-1}|b^{i}|\\
    &= |b^n| \frac{1-|b^k|}{1-|b|}\\
    &< |b^n|\frac{1}{1-|b|} \text{  , since } 0<|b|<1.
\end{align*}
Given an $ \epsilon>0 $ we need to find $ N \in \mathbb{N}$ 
such that for $ m,n>N $,  $ |a_m-a_n|<\epsilon $. 
We want $ |a_m-a_n| <|b^n| \frac{1}{1-|b|}< \epsilon$. 
$$ |b^n| < \epsilon (1-|b|) $$
$$ n \log(|b|) < \log({\epsilon (1-|b|)}) $$
And since $ \log(|b|) <0$ since $ |b|<1<e  $ it follows that 
$$ n>\frac{\log(\epsilon(1-|b|))}{\log(|b|)} .$$
So given any $ \epsilon>0 $, for any $n,m> \frac{\log(\epsilon(1-|b|))}{\log(|b|)}$ it follows that $ |a_{m}-a_{n}| < \epsilon $. Therefore $  \{a_n\}$ is Cauchy.
My issue with this is that we have yet to cover functions such as $\log$. I rely on some properties of logs to solve this since we have an exponent to deal with. Is there a way to get around this?
My issue with this is that we have yet to cover functions such as $\log$. I rely on some properties of logs to solve this since we have an exponent to deal with. Is there a way to get around this?
 A: You don't need to express an $N$ explicitly: the $b^n \to 0$ will guarantee the existence of such an $N$, and that's enough.
Edit:
$|b^n| \to 0$ means that $\forall \varepsilon_0 > \exists N$ so that $n > N \implies |b^n| < \varepsilon_0$. Let $\varepsilon_0=(1-|b|)\varepsilon$. We can do it, because it's greater than $0$. So we will have that
$$|b^n|\frac{1}{1-|b|} < \frac{\varepsilon_0}{1-|b|}=\frac{(1-|b|)\varepsilon}{1-|b|}=\varepsilon$$
for all $n>N$, as we desired.
A: You were on the right track to get an easier answer, at least if you are familiar with basis series.
Let $q> p$. Then
$$|a_p-a_q| \leq \sum_{n=p+1}^q |a_n-a_{n-1}|\leq \sum_{n={p+1}}^q b^n $$
Now, notice that $\sum_{n=1}^\infty b^n$ is a convergent series (because $b < 1)$. In particular, the sequence of partial sums $\left(\sum_{k=1}^n b^k\right)_{n=1}^\infty$ is a Cauchy sequence. This allows us to make the sum $\sum_{n={p+1}}^q b^n $ as small as we desire, since $\sum_{n={p+1}}^q b^n  = \sum_{n=1}^q b^n - \sum_{n=1}^p b^n$.
A: Since $\frac1b>1$, you have, by Bernoulli's inequality, that$$\frac1{b^n}=\left(\frac1b\right)^n\geqslant1+n\left(\frac1b-1\right)>n\left(\frac1b-1\right)$$and therefore$$b^n<\frac1{n\left(\frac1b-1\right)}.$$So, in order to have $b^n<\varepsilon(1-b)$, all you need is that $\frac1n<\varepsilon(1-b)\left(\frac1b-1\right)$. No logarithms are need here.
A: You can first prove that $\lim_n |b|^n=0$. This can be done without invoking the $\log$ function. If $b=0$, we are done, otherwise, since $1/|b|>1$, there is a positive $h$ such that, 
$$\dfrac{1}{|b|}=1+h$$
and, by Bernoulli's inequality,
$$\dfrac{1}{|b|^n}=(1+h)^n\geqslant1+nh.$$
Therefore, 
$$0\leqslant |b|^n\leqslant \dfrac{1}{1+nh},$$
and, since $\lim_n1/(1+nh)=0$, by Squeezing Theorem, $\lim_n|b|^n=0$. 
Now let $\epsilon>0$ be given. Then
\begin{align*}
\lim_n|b|^n=0\Rightarrow \exists N\in\mathbb{N}\;\forall n\;\big(n\geqslant N\Rightarrow |b|^n<\epsilon(1-|b|)\big).
\end{align*}
Hence, for natural numbers $m$ and $n$ with $m>n\geqslant N$, as you have showed,
$$|a_m-a_n|<|b|^n\dfrac{1}{1-|b|},$$
so
$$|a_m-a_n|<\epsilon.$$
A: More generally,
if $b_n > 0$
and $\sum b_n$ converges,
then
$|a_{n+1}-a_n| \le b_n
$
implies that
$(a_n)_{n=1}^{\infty}
$
is Cauchy.
You just have to use 
the definition of convergent
and the triangle inequality.
