Working with Geometric Series BC only

Multiplying $S_n$ by $r$ and subtracting gives the closed form $S_n = a\frac{1-r^{n}}{1-r}$, and since $r^{n}$ is the only $n$-dependent piece, the limit exists precisely when $|r| < 1$, giving $\frac{a}{1-r}$. Both endpoints fail: $r = 1$ makes $S_n = na$, and $r = -1$ makes the partial sums alternate between $a$ and $0$ without settling, so the condition is strict. The formula therefore describes convergent geometric series rather than geometric series in general, and applying it to $\sum 2^{n}$ returns $-1$ for a sum of positive terms.

The numerator is the first term actually present, not the coefficient out front: $\sum_{n=2}^{\infty} 3\left(\frac{1}{2}\right)^{n}$ has $a = \frac{3}{4}$ and sums to $\frac{3}{2}$, while the same three symbols summed from $n = 0$ or $n = 1$ give $6$ and $3$, and those three check against each other since $6 - 3 - \frac{3}{2} = \frac{3}{2}$. The ratio is safest computed as $\frac{a_{n+1}}{a_n}$, which survives an exponent of $2n$ (where $r = \frac{4}{9}$, not $\frac{2}{3}$), a sign that belongs inside $r$, and a $2^{n}$ in a denominator that makes $r = \frac{1}{2}$ rather than $2$. The closed form also evaluates every repeating decimal, since $0.\overline{7} = \frac{7/10}{9/10} = \frac{7}{9}$, and supplies the first stock of known series for the comparison tests of 10.6.