In my econometric book, it says
$\sum\limits_{i=1}^n a^{i-1}=(1-a^n)/(1-a)$ because $\sum\limits_{i=1}^n a^{i-1}=(1-a^n)\sum\limits_{i=0}^\infty a^i$.
I understand the proof of $\sum\limits_{i=0}^\infty a^i=1/(1-a)$ and $\sum\limits_{i=1}^n a^{i-1}=(1-a^n)/(1-a)$ separately. But how do we know $\sum\limits_{i=1}^n a^{i-1}=(1-a^n)\sum\limits_{i=0}^\infty a^i$, assuming we don't know the representation of $\sum\limits_{i=0}^\infty a^i$ and $\sum\limits_{i=1}^n a^{i-1}$?
It should be true only for $|a|<1$ since
$$\sum\limits_{i=1}^n a^{i-1}=\frac{1-a^n}{1-a} \implies \sum\limits_{i=0}^\infty a^i=\frac1{1-a}$$
and therefore
$$\sum\limits_{i=1}^n a^{i-1}=(1-a^n)\sum\limits_{i=0}^\infty a^i=\frac{1-a^n}{1-a}$$
otherwise the equality doesn’t hold since the series doesn’t converge.