Which $a$, which $r$, and does it converge BC only

For $\sum_{n=0}^{\infty} ar^{n}$ the partial sum has a closed form. Multiply $S_n = a + ar + \cdots + ar^{n-1}$ by $r$, subtract, and everything in the middle cancels:

$$S_n = a\,\frac{1 - r^{n}}{1 - r}, \qquad r \neq 1.$$

Now take the limit, which is where the condition appears. The only part of that expression depending on $n$ is $r^{n}$, so:

1. If $|r| < 1$: $r^{n} \to 0$, so $S_n \to \dfrac{a}{1-r}$. The series converges.
2. If $|r| > 1$: $r^{n}$ grows without bound and so does $S_n$. Divergent.
3. If $r = 1$: the series is $a + a + a + \cdots$, and $S_n = na$. Divergent for $a \neq 0$.
4. If $r = -1$: the partial sums alternate between $a$ and $0$ and never settle. Divergent.

Both endpoints are excluded, so the condition is the strict inequality $|r| < 1$. The formula is not a fact about geometric series; it is a fact about convergent ones. Applying it to $\sum 2^{n}$ returns $\frac{1}{1-2} = -1$, a negative number for a sum of positive terms, which is the formula answering a question it was never asked.

Written $\sum_{n=0}^{\infty} ar^{n}$, the first term is $ar^{0} = a$ and the numerator looks like a coefficient. That coincidence is the trap, because it stops being true the moment the index moves.

Consider $\displaystyle\sum_{n=2}^{\infty} 3\left(\frac{1}{2}\right)^{n}$. The coefficient is $3$. The first term actually present is the $n = 2$ term:

$$a = 3\left(\frac{1}{2}\right)^{2} = \frac{3}{4}, \qquad \text{so} \qquad S = \frac{3/4}{1 - 1/2} = \frac{3}{2}.$$

Writing $\frac{3}{1 - 1/2} = 6$ instead is not a slip in algebra. It is the correct sum of a different series, the one starting at $n = 0$. The three versions of this series are all defensible answers to questions nobody asked:

1. From $n = 0$: first term $3$, sum $6$.
2. From $n = 1$: first term $\frac{3}{2}$, sum $3$.
3. From $n = 2$: first term $\frac{3}{4}$, sum $\frac{3}{2}$.

They check against each other: $6 - 3 - \frac{3}{2} = \frac{3}{2}$, since dropping the $n=0$ and $n=1$ terms from the full sum must leave the third. The reliable habit is to substitute the lower limit and compute the first term, every time, rather than reading a coefficient off the front.

The ratio is what each term is multiplied by to produce the next, so the definition itself is the safest way to find it:

$$r = \frac{a_{n+1}}{a_n}.$$

Compute that quotient once and simplify. It survives every disguise the notation can put on a series, and three of those are worth naming:

1. An exponent that is not $n$. In $\sum 5\left(\frac{2}{3}\right)^{2n}$ the ratio is $\left(\frac{2}{3}\right)^{2} = \frac{4}{9}$, not $\frac{2}{3}$, because advancing $n$ by one advances the exponent by two.
2. A negative ratio. $\sum 4(-1)^{n}3^{-n}$ has $r = -\frac{1}{3}$. The sign belongs in $r$, and $\left|-\frac{1}{3}\right| < 1$ so it converges, to $\frac{4}{1+1/3} = 3$.
3. A ratio written upside down. $\sum \frac{3}{2^{n}}$ has $r = \frac{1}{2}$, not $2$. The $2$ is visible and the $\frac{1}{2}$ is the thing being multiplied.

An inverted ratio is the most damaging of the three, because it can flip a convergent series into an apparently divergent one or produce a negative sum for positive terms. Both are signals worth acting on.

Two uses beyond evaluation, both of which come back later in the unit.

Repeating decimals. $0.\overline{7}$ is $\frac{7}{10} + \frac{7}{100} + \cdots$, a geometric series with $a = \frac{7}{10}$ and $r = \frac{1}{10}$, so it equals $\frac{7/10}{9/10} = \frac{7}{9}$. Every repeating decimal is a geometric series, which is why every one of them is rational.

A benchmark for the comparison tests. From 10.6 onward you will need series whose behaviour is already known, and the geometric family is the first such stock. A series that eventually behaves like $\left(\frac{1}{2}\right)^{n}$ converges; one that behaves like $2^{n}$ does not.

Finally, note what the closed form is not. It gives the sum only when the series converges, and no other series in this course has one. From 10.3 onward the question stops being "what is the sum" and becomes "does a sum exist", because for almost every series the second question is answerable and the first is not.