AP Calculus BC

When a Geometric Series Diverges

A geometric series converges exactly when the common ratio is less than one in absolute value, and the sum formula is only valid in that case. At a ratio of one the terms never shrink, the partial sums climb without bound, and the formula would divide by zero.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1. 1

   The common ratio satisfies |r| < 1

   $$|r| < 1$$

   The whole behaviour turns on whether $r^n$ shrinks. When it does, the partial sums close in on a finite value and the formula follows from the finite sum. When it does not, the terms fail to vanish and the nth term test finishes the argument immediately.

   Drop it and the theorem fails

   $$\sum_{n=0}^{\infty} 1 \cdot 1^{n} = 1 + 1 + 1 + \cdots$$

   The geometric series with common ratio 1

   With $r = 1$ every term is 1, so the terms never tend to zero and the check confirms they are still 1 at the millionth index. The series diverges by the nth term test. Notice also that the formula $a/(1-r)$ would ask you to divide by zero here, which is the algebra warning you about the same thing.

2. 2

   The index starts where the formula assumes it starts

   $$\sum_{n=0}^{\infty} ar^{n}, \text{ not } \sum_{n=1}^{\infty}$$

   The formula $a/(1-r)$ takes a as the FIRST term actually present. Starting at $n = 1$ instead makes the first term $ar$, so the sum is $ar/(1-r)$. Nothing about convergence changes, since dropping finitely many terms never changes whether a series converges, but the value does. This is a bookkeeping hypothesis rather than an analytic one, so no counterexample is displayed; the safe habit is to read off the first term you can see and divide by one minus the ratio.

   No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

1. Write the partial sum $S_N = a + ar + \cdots + ar^{N-1}$ and subtract $rS_N$ from it. Everything cancels except the ends.
2. That gives $S_N(1 - r) = a(1 - r^N)$, so $S_N = a\frac{1 - r^{N}}{1 - r}$ whenever $r \ne 1$.
3. If $|r| < 1$ then $r^N \to 0$, so $S_N \to a/(1-r)$ and the series converges to that value.
4. If $|r| \ge 1$ then $r^N$ does not tend to 0, so neither do the terms, and the nth term test gives divergence. The case $r = 1$ has to be handled separately, since the algebra above divided by $1 - r$.

What it does not say