I was trying to solve a simple probability problem I found in a book. The procedure was wrong, but, by chance, I found that if $a = 2$ and $n = 6$ (my wrong solution to the problem) $ \sum_{i=1}^6 (1/2)^i = 1-\frac{1}{2^6} $. I tried to make a proof and I write here:
$$S_{n} = \sum_{i=1}^n \frac{1}{2^i} = \frac{1}{2} + \frac{1}{2^2} +...+ \frac{1}{2^n}$$
$$2S_{n} = 2\frac{1}{2} + 2\frac{1}{2^2} +...+ 2\frac{1}{2^n}$$
$$2S_{n} = 1 + \sum_{i=1}^{n-1}\frac{1}{2^i}$$
$$2S_{n} = 1 + \sum_{i=1}^{n-1}\frac{1}{2^i} - \sum_{i=1}^n \frac{1}{2^i}$$
$$S_{n} = \sum_{i=1}^n \frac{1}{2^i} = 1 - \frac{1}{2^n}$$
Latter, I wonder if that would work similarly to other fractions and I found the formula presented in the question, which applies for any value of $a$. An which proof is as follows:
$$S_{n} = \sum_{i=1}^n \frac{1}{a^i} = \frac{1}{a} + \frac{1}{a^2} +...+ \frac{1}{a^n}$$
$$(a-1)S_{n} = \frac{a-1}{a} + \frac{a-1}{a^2} +...+ \frac{a-1}{a^n}$$
$$(a-1)S_{n} = 1 - \frac{1}{a} + \frac{1}{a} - \frac{1}{a^2} +...+ \frac{1}{a^{n-1}} - \frac{1}{a^n}$$
$$(a-1)S_{n} = 1 - \frac{1}{a^n}$$
$$\sum_{i=1}^n \frac{1}{a^i} = \frac{1 - \frac{1}{a^n}}{a-1}$$
When the limit is taken as $n \to \infty$ approaches infinitiy, the result is the same as that of the infinite sum of a geometric series.
I suppose this is not new, but since it was something I did out of fun and discover it mysefl (I was actually very happy to see the answer), I wonder the applications it has. Or any real world situacion when it is useful. I am not a mathematician (veeeery far from being one) and I didn't find anything related other than the result of a sum of an infinite geometric series, which is found differently.
Best regards,
Sebastián