Sum to Infinity

An infinite GP has a finite total exactly when its terms shrink fast enough:

$S_\infty = \dfrac{a}{1 - r}$, valid only when $|r| < 1$

The intuition: $S_n = \dfrac{a(1 - r^n)}{1 - r}$, and when $|r| < 1$ the $r^n$ term dies as $n$ grows, leaving $\dfrac{a}{1 - r}$. That one-line derivation is worth knowing. “explain why the sum to infinity exists” wants the condition and the dying-$r^n$ idea.

The condition is not decoration

$|r| < 1$ is part of every answer in this subtopic. Three ways 0606 cashes it:

• “Find $S_\infty$”, check $|r| < 1$ before computing; if $r = 3$, the correct answer is ”$S_\infty$ does not exist, since $|r| = 3 > 1$”, the statement is the mark
• “Explain why the sum to infinity exists”, quote the condition with your $r$
• “Find the values of $x$ for which the GP has a sum to infinity”, the ratio contains $x$; solve $|r(x)| < 1$, i.e. $-1 < r(x) < 1$, a sandwich inequality. This is the modern exam’s favourite dress for the condition.

GP: $8$, $6$, $4.5$, … Find $S_\infty$. $r = \tfrac{6}{8} = 0.75$; $|r| < 1$ ✓ (state it) $S_\infty = \dfrac{8}{1 - 0.75} =$ $32$

Reverse problems

Given $S_\infty$ and one other fact, recover $a$ and $r$:

$S_\infty = 20$ and $a = 5$: $\dfrac{5}{1 - r} = 20 \to 1 - r = \tfrac{1}{4} \to$ $r = \tfrac{3}{4}$

With two symbolic facts (e.g. $S_\infty = 27$ and $u_2 = 6$), translate each into an equation and solve simultaneously, expect a quadratic in $r$ and apply the $|r| < 1$ filter to its roots, in writing. The filter frequently eliminates exactly one root: that’s the design.

A related regular: “find the least $n$ for which $S_n$ exceeds 99% of $S_\infty$”, set up $\dfrac{S_n}{S_\infty} = 1 - r^n > 0.99$, solve with logs (mind the inequality flip when dividing by the negative log of $r < 1$), answer with an integer $n$.

Common mistakes

• $S_\infty$ computed for $|r| \ge 1$
• The condition checked mentally but never written
• “Values of $x$” questions answered with $r(x) < 1$ only (the left half, $r(x) > -1$, forgotten)
• $1 - r$ inverted ($S_\infty = \dfrac{a}{r - 1}$ sign error)
• The 99%-of-$S_\infty$ log inequality flipped the wrong way

Full topic context: Series notes · prerequisite: geometric progressions.