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.
 A: As @TheoBendit suggested, comparison with a telescoping series can work:
For example, the classic telescoping series
$$
\sum_{0\le n\le N} {1\over n(n+1)}
\;=\; 
\sum_{0\le n\le N} \Big({1\over n}-{1\over n+1}\Big)
\;=\;
{1\over 1} - {1\over N+1}
$$
Similarly, the tails go to zero, so by whatever criterion we like, this converges.
On the other hand, for any $|C|<1$, for large-enough $n$, $|C^n|\le {1\over n(n+1}$. So, disregarding finitely-many terms, the telescoping series dominates the (convergent) geometric series.
A: For $C \in [0,1)\,$ the sequence is monotonically increasing and bounded above, thus convergent.

*

*Increasing: $\;s_n = \sum_{k=0}^n C^k = s_{n-1} +C^n \ge s_{n-1}\,$.


*Bounded: $\;s_n= 1 + C\,s_{n-1}\,$, then assuming $s_{n-1} \le L \implies s_n \le 1 + C\,L$. The latter implies $s_n \le L$ when $1 + C\,L \le L \iff L \ge \frac{1}{1-C}\,$, so in particular $\frac{1}{1-C}$ is an upper bound.
When $C \in (-1,0)$ the partial sum $\sum_{0\leq k \leq n} C^k = \sum_{0\leq 2k \leq n} \left(C^2\right)^k + C\sum_{0\leq 2k + 1 \leq n} \left(C^2\right)^k$ where $C^2 \in (0,1)$ so the problem reduces to the first case.
A: If you can show it for a single $C$ (ex:$1/2$ using Zeno’s argument), then you can choose $n$ so that $c^n<1/2$, break the sequence into $n$ sequences of every $n$th value. Each of these sequences converges by comparing with the $1/2$ sequence. There’s a finite number of them, so by interleaving them, the whole sequence converges.
A: Natural Derivation of the Closed Form
The first way of handling geometric series I saw, was to derive the closed form for the finite sum. Discovering this derivation seems pretty reasonable if one stares at the geometric sum long enough:
$$
\begin{align}
S&=1+a+a^2+\dots+a^{n-1}\tag{1a}\\
aS&=\phantom{1+{}}a+a^2+\dots+a^{n-1}+a^n\tag{1b}\\
(1-a)S&=1\phantom{{}+a+a^2+\dots+a^{n-1}}-a^n\tag{1c}\\
S&=\left.\left(1-a^n\right)\middle/(1-a)\right.\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: write out the summation
$\text{(1b)}$: multiply by $a$
$\text{(1c)}$: subtract $\text{(1b)}$ from $\text{(1a)}$
$\text{(1d)}$: divide by $1-a$

Euler-Maclaurin Sum Formula
The Euler-Maclaurin Sum Formula can also be applied. If we ignore the error term, the Euler-Maclaurin Sum Formula becomes $\frac{D}{1-e^{-D}}D^{-1}f$, where $D^{-1}$ is indefinite integration (which introduces the constant that appears in the formula). This is discussed a bit in this answer.
Let $a\gt0$ and $a\ne1$. Note that $D^{-1}a^x=\frac{a^x}{\log(a)}+C$. For non-negative $k\in\mathbb{Z}$, $D^ka^x=\log(a)^ka^x$.
$$
\begin{align}
\frac{D}{1-e^{-D}}D^{-1}a^x
&=\frac{D}{1-e^{-D}}\left(\frac{a^x}{\log(a)}+C\right)\tag{2a}\\[3pt]
&=\frac{\log(a)}{1-\frac1a}\frac{a^x}{\log(a)}+C\tag{2b}\\
&=\frac{a^{x+1}-1}{a-1}\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply $D^{-1}$
$\text{(2b)}$: each application of $D$ simply introduces a factor of $\log(a)$
$\phantom{\text{(2b):}}$ thus, an application of $f(D)$ simply multiplies by $f(\log(a))$
$\text{(2c)}$: simplify and set $C=-\frac1{a-1}$ to match $\sum\limits_{k=0}^xa^k$ at $x=0$
For $a\in\left(e^{-2\pi},e^{2\pi}\right)\setminus\{1\}$, the series given by the Euler-Maclaurin Sum Formula actually converges, rather than merely giving an asymptotic expansion.
I am not proposing the Euler-Maclaurin Sum Formula as a way to initially approach the Geometric Sum Formula. I present it more as a point of interest. A demonstration of analyzing the series as one might other, more complicated, series.
A: 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}
$
