What is the sum of the **infinite** geometric series $54 + 18 + 6 + 2 + \dfrac{2}{3} + \dfrac{2}{9} + \ldots$?

First confirm the sequence is geometric and identify the ratio: $r = \dfrac{18}{54} = \dfrac{1}{3}$. Verify: $\dfrac{6}{18} = \dfrac{1}{3}$ ✓ and $\dfrac{2}{6} = \dfrac{1}{3}$ ✓. Because $|r| = 1/3 < 1$, the infinite sum **converges** and equals $S_{\infty} = \dfrac{a_{1}}{1 - r} = \dfrac{54}{1 - 1/3} = \dfrac{54}{2/3} = 54 \cdot \dfrac{3}{2} = 81$. - Trap A ($54$) returns just the first term. - Trap B ($72$) is close to a partial sum after a few terms but not the limit. - Trap D would apply if $|r| \ge 1$; here $|r| = 1/3$, so the series converges. The magic of $|r| < 1$: the formula $\dfrac{a_{1}}{1 - r}$ collapses an infinite sum to a single quotient. For $r = 1/3$, the multiplier is $\dfrac{1}{1 - 1/3} = \dfrac{3}{2}$, so the total ends up $1.5\times$ the first term.