Starting index for geometric series test

Question

I was just wondering if the geometric series test for series of the form $\sum_{n=}^{\infty}ar^{n}$ needs the index to start at $0$ or $1$? From my understanding of the proof using partial sums, calculating the convergent value as $\frac{a}{1-r}$ requires the series to start at $0$. I ask because I've been seeing a lot of post and even my assignment solutions neglecting whether it starts at $1$ or $0$. If my assignment assumes I do not know how to change indices when it starts at $n=1$, and this happens to not be negligible, how do I go around using the geometric series test and calculating the convergent value?

Answer 1

Recall that for the geometric series

$$\ S_n = \sum_{j=0}^n ar^j = \frac{ar^{n+1}-a}{r-1}\implies \quad |r|<1\quad S_{\infty}=\frac{a}{1-r} $$

then if we start from $n=1$

$$\ S_n = \sum_{j=1}^n ar^j = \frac{ar^{n+1}-a}{r-1}-a\implies \quad |r|<1\quad S_{\infty}=\frac{a}{1-r}-a$$

Answer 2

The starting index is irrelevant to determine whether a geometric series converges (or in general whether a series converges). All that matters to see if a geometric series converges is that the common ratio $r$ be such that $|r|<1$ . Further the value of a geometric series with initial term $a$ and common ratio $r$ is $$ \frac{a}{1-r} $$