Alternative proofs of convergence of geometric series The usual proof for the convergence of a geometric series of ratio $C: |C|\in [0,1)$ makes use of the formula $$\sum_{0\leq k \leq n} C^k = \frac{1-C^{n+1}}{1-C}.$$
I'm looking for alternative ways to prove it. The motivation for this is that, if someone who never saw this formula tried to prove the geometric series converges might have a hard time, unless maybe there are other, perhaps more insightful ways to prove it.

If $c = \dfrac1{1+b}$ where $b > 0$
then,
since,
for $k \ge 2$,
$(1+b)^k
\ge 1+bk+b^2k(k-1)/2
\gt b^2k(k-1)/2$
(readily proved by induction),
so
$c^k 
= \dfrac1{(1+b)^k}
\le \dfrac{2}{b^2k(k-1)}
$
so
$\begin{array}\\
\sum_{k=2}^n c^k 
&\le \sum_{k=2}^n\dfrac{2}{b^2k(k-1)}\\
&= \dfrac{2}{b^2}\sum_{k=2}^n\dfrac1{k(k-1)}\\
&= \dfrac{2}{b^2}\sum_{k=2}^n(\dfrac1{k-1}-\dfrac1{k})\\
&= \dfrac{2}{b^2}(1-\dfrac1{n})\\
&< \dfrac{2}{b^2}\\
&= \dfrac{2}{(\frac1{c}-1)^2}\\
&= \dfrac{2c^2}{(1-c)^2}\\
&\text{so}\\
\sum_{k=0}^n c^k 
&< 1+c+\dfrac{2c^2}{(1-c)^2}\\
&= \dfrac{c^3 + c^2 - c + 1}{(1-c)^2}\\
\end{array}
$
