Infinite Geometric Series Formula Derivation We know that the formula for computing a geometric series is:$$\sum_{i=1}^{\infty}{a_0r^{i-1}} = \frac{a_0}{1-r}$$
Out of curiosity, I would like ask: Is there any ways the formula can be derived other than the following two ways?

Method 1 (The way I found on my own):
$$\sum_{i=1}^{\infty}{a_0r^{i-1}} \equiv S$$
$$S = a_0r^0+a_0r^1+a_0r^2+\cdots$$
$$S = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots\right)$$
$$S = r\left(a_0r^{-1} + S\right)$$
$$S = a_0 + rS$$
$$(1-r)S = a_0$$
$$S = \frac{a_0}{(1-r)}$$
Note that for this to work, you must first confirm this: $$\lim_{n\to\infty} a_n = 0$$
Method 2 (The way I found on the web):
$$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$
$$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}$$
$$rS_n = r\left(a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}\right)$$
$$rS_n = a_0r^1 + a_0r^2 + a_0r^3 + \cdots + a_0 r^{n-1} + a_0 r^{n}$$
$$S_n-rS_n = a_0r^0 - a_0r^n$$
$$(1-r)S_n = a_0 - a_0 r^n$$
$$S_n = \frac{a_0(1 - r^n)}{1-r}$$
Given:$$\left|r\right| < 1,$$
$$\lim_{n\to \infty} S_n = \lim_{n\to \infty}\frac{a_0(1 - r^n)}{1-r} = \frac{a_0}{1-r}$$
I personally prefer Method 1 because it is faster and more intuitive, as we don't have to multiply by $r$.

Method 1 for formula of partial sums:
$$\sum_{i=1}^{n}{a_0r^{i-1}} \equiv S_n$$
$$S_n = a_0r^0+a_0r^1+a_0r^2+\cdots+a_0r^{n-2}+a_0r^{n-1}$$
$$S_n = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots+a_0r^{n-3}+a_0r^{n-2}\right)$$
$$S_n = r\left(a_0r^{-1} + S_n - a_0r^{n-1}\right)$$
$$S_n = a_0 + rS_n - a_0r^{n}$$
$$(1-r)S_n = a_0 - a_0r^n$$
$$S_n = \frac{a_0(1 - r^n)}{(1-r)}$$

$a+ar+a(r^{2})...=s$
$a+rs=s$
$a+rs-s=0$
$a+(r-1)s=0$
$-a=(r-1)s$
$\frac{-a}{r-1}=s$
$r-1=-(1-r)$
$\frac{-a}{-(1-r)}=s$
$s=\frac{a}{1-r}$
