Proof verification. $\{x_n\}$ is a sequence such that $|x_{n+1} - x_n| \le C\alpha^n$ for $\alpha\in (0, 1), n\in\Bbb N$. Prove $x_n$ converges. 
Let $\{x_n\}, n\in \Bbb N$ denote a sequence such that:
  $$
\begin{cases}
|x_{n+1} - x_n| \le C\alpha^n \\
0 < \alpha < 1
\end{cases}
$$
  Prove $\{x_n\}$ converges.

Given the fact $|x_{n+1} - x_n| \le C\alpha^n$ consider the following inequalities:
$$
|x_{n+1} - x_n| \le C\alpha^n \\
|x_{n+2} - x_{n+1}| \le C\alpha^{n+1} \\
|x_{n+3} - x_{n+2}| \le C\alpha^{n+2} \\
\dots \\
|x_{n+p+1} - x_{n+p}| \le C\alpha^{n+p} \\
$$
Consider the sum of the inequalities:
$$
|x_{n+1} - x_{n}| + |x_{n+2} - x_{n+1}| + |x_{n+3} - x_{n+2}| + \cdots + |x_{n+p+1} - x_{n+p}| \\ 
\le \sum_{k=0}^p C\alpha^{n+k} = C \sum_{k=0}^p \alpha^{n+k} \tag1
$$
By geometric sum:
$$
C \sum_{k=0}^p \alpha^{n+k} = C \cdot \frac{\alpha^n(1-\alpha^{p + 1})}{1-\alpha} \le C \cdot \frac{\alpha^n}{1-\alpha}
$$
Lets fix some $\epsilon > 0$, and $N\in \Bbb N$ such that:
$$
C\cdot \frac{\alpha^{N}}{1-\alpha} < \epsilon
$$
Rewrite $\alpha$ as:
$$
\alpha = \frac{1}{1+r},\ r \in \Bbb R_{>0}
$$
Thus:
$$
C\cdot \frac{1}{(1-\alpha)(1+r)^N} < \epsilon \\
(1+r)^N > {C\over (1-\alpha)\epsilon} \\
N > \log_{1+r} {C\over (1-\alpha)\epsilon}
$$
Returning to $(1)$ we have by triangular inequality:
$$
|x_{n}- x_{n+1} + x_{n+1} - x_{n+2} + \cdots + x_{n+p} - x_{n+p+1}| \\ \le |x_{n+1} - x_{n}| + |x_{n+2} - x_{n+1}| + \cdots + |x_{n+p+1} - x_{n+p}|
$$
Since the values are telescoping we obtain:
$$
|x_{n} - x_{n+p+1}| < C \cdot \frac{\alpha^n}{1-\alpha} < \epsilon
$$
Now if we choose:
$$
n > N > \log_{1+r} {C\over (1-\alpha)\epsilon}
$$
we obtain a regular definition of the Cauchy criteria, which means $\{x_n\}$ is a convergent sequence.
Could you please verify the reasoning above? Thank you!
 A: Seems fine. 
We have $C \ge 0$. 
In the event that $C=0$, we have a constant sequence and hence it converges. 
A: This is a comment
rather than an answer.
You don't have to use logs.
By Bernoulli's inequality,
$(1+r)^N \ge 1+rN
\gt rN$
so that
$C \frac{1}{(1-\alpha)(1+r)^N}
\lt C\frac{1}{(1-\alpha)rN}
$.
Therefore,
if
$\epsilon
\gt  C \frac{1}{(1-\alpha)rN}
$,
or
$N >  \frac{C}{(1-\alpha)r\epsilon}
$,
then
$C \frac{1}{(1-\alpha)(1+r)^N} < \epsilon
$.
Of course 
this isn't as good
as the equation using logs,
but it is completely elementary.
A: You have the right idea but it could be briefer: Let $A(n)=\sup_{m\in \Bbb N}|x_n-x_{n+m}|.$ The Cauchy Criterion for convergence of $(x_n)_{n\in \Bbb N}$ is $\lim_{n\to \infty}A(n)=0.$
Since $0<\alpha<1$ we have $\sup_{m\in \Bbb N}(1-\alpha^m)=1,$ so  $$A(n)=\sup_{m\in \Bbb N}|x_n-x_{n+m}|\leq$$ $$\leq \sup_{m\in \Bbb N}C\alpha^n\cdot\frac {1-\alpha^m}{1-\alpha}=$$ $$=C\alpha^n\cdot \frac {1}{1-\alpha}.$$ This last expression goes to $0$ as $n\to \infty$ because $0<\alpha <1$.
