If $c\in\mathbb R$, $0 < r < 1$, and $|s_{n+1} - s_n|\le cr^n$ for all $n\in\mathbb N$, show the sequence of $\mathbb R$, $\{s_n\}$ converges 
If $c\in\mathbb R$, $0 < r < 1$, and $|s_{n+1} - s_n|\le cr^n$ for all $n\in\mathbb N$, show the sequence of $\mathbb R$, $\{s_n\}$ converges

I thought that perhaps we might be able to show that the sequence is Cauchy because under $\mathbb R$, we know that Cauchy implies convergence. Here's what I've done so far
$$\begin{align}|s_m - s_n| &= |s_m - s_{m-1} + s_{m-1} - s_{m-2} + s_{m-2} + \cdots + s_{n+1} - s_n|\\ &\le |s_m - s_{m-1}| + |s_{m-1} - s_{m-2}| + |s_{m-2} - s_{m-3}| + \cdots + |s_{n+1} - s_n|\\ &\le cr^{m-1} + cr^{m-2} + \cdots + cr^n \\ &= c\sum_{i = n}^{m-1}r^i\\&\text{I'm stuck here though. }\\ &\text{I think $\sum_{i = n}^{m-1}r^i = {r^m + r^n \over 1- r}$, so:}\\& = c\left({r^m + r^n \over 1- r}\right) \\ &\le c\left({r^n + r^n \over 1-r}\right) \\ &= 2c\left({r^n \over 1- r}\right) < \epsilon\end{align}$$
I'm not sure how to proceed from here, or if I even did the summation correct. I said $$s_n = cr^{m-1} + cr^{m-2} + \cdots + cr^n$$ and $$rs_n = cr^m + cr^{m-1} + \cdots + cr^{n+1},$$ so $$s_n-rs_n = cr^m + cr^n \implies s_n = c\left({r^m+r^n \over 1 - r}\right)$$ and additionally on the assumption that $n \le m$, we have that $r^n \ge r^m$ since $0 < r < 1$.
But I'm not guaranteed anything about $c$ other than it is nonnegative since again $0<r<1$. Thus, how can I choose my $N$ such that for all $m\ge n\ge N$, $|s_m - s_n| < \epsilon$ for all $\epsilon > 0$? I tried observing $r^n < {\epsilon(1-r) \over 2c}$ from the last step, but what happens when $c = 0$?
 A: I think on the whole you are on the right track. There are just a few holes to fill in. 
Disclaimer: I wrote this very sleepy so let me know if I messed up anywhere :)

If $c\in\mathbb R$, $0 < r < 1$, and $|s_{n+1} - s_n|\le cr^n$ for all $n\in\mathbb N$, show the sequence of $\mathbb R$, $\{s_n\}$ converges

Let's begin with what you had. We have that for all $m,n$ assuming $m \geq n$:
$$\begin{align}|s_m - s_n| &= |s_m - s_{m-1} + s_{m-1} - s_{m-2} + s_{m-2} + \ldots + s_{n+1} - s_n|\\ &\le |s_m - s_{m-1}| + |s_{m-1} - s_{m-2}| + \ldots + |s_{n+1} - s_n|\\ &\le cr^{m-1} + cr^{m-2} + \ldots + cr^n \\ &= c\sum_{i = n}^{m-1}r^i\\
& = c\left( \frac{r^n - r^m}{1-r}\right) \\
& \tag{1} < c\left( \frac{r^n}{1-r}\right) \\
\end{align}
$$
So let us prove the convergence:
Let $\epsilon > 0$. We want to prove there exists and $N$ such that $\forall m,n > N$, we have $\vert s_m-s_n \vert < \epsilon$.
Let $N = \left\lceil{\frac{\ln{\left(\frac{\epsilon(1-r)}{c}\right)}}{\ln(r)}}\right\rceil$ or any $N$ that satisfies $c\left( \frac{r^N}{1-r}\right) < \epsilon$.
Note: we are assuming $c > 0$ since if $c = 0$, the proof is trivial.
Then we have:
$$
c\left( \frac{r^N}{1-r}\right) < \epsilon
$$
Notice for all $n > N$, the following holds:
$$
\tag{2}
c\left( \frac{r^n}{1-r}\right) < c\left( \frac{r^N}{1-r}\right) < \epsilon 
$$.
So consider any $m,n > N$ where WLOG $m \geq n$and consider $\vert s_m - s_n \vert$. From (1), we have that:
$$
|s_m - s_n| < c\left( \frac{r^n}{1-r}\right) \\
$$
and since $n > N$, from (2) we have:
$$
|s_m - s_n| < c\left( \frac{r^n}{1-r}\right) < c\left( \frac{r^N}{1-r}\right) < \epsilon \\
\implies |s_m - s_n| < \epsilon
$$
as required.
