Why Does a Geometric Series Only Converge When |r| < 1?

A non-zero infinite geometric series converges only when $\lvert r\rvert<1$ because successive terms then shrink towards zero, causing the sequence of partial sums to approach a finite limit. If $\lvert r\rvert\geq 1$, the partial sums do not settle at one finite value.

The Mechanism

For first term $a$ and common ratio $r$, the sum of the first $n$ terms is

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

An infinite series converges if the partial sums $S_n$ approach a finite value as $n\to\infty$. This requires $r^n\to 0$, which happens precisely when .

The condition that individual terms approach zero is necessary but, for a general series, it is not sufficient for convergence. For a geometric series, however, the partial-sum formula shows that the behaviour of $r^n$ completely determines convergence. Each new term changes the running total by a smaller amount, so the partial sums close in on a fixed number.

Therefore,

$$S_\infty=\lim_{n\to\infty}\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}.$$

For example,

$$8+4+2+1+\cdots$$

has $a=8$ and $r=\frac{1}{2}$. Therefore,

$$S_\infty=\frac{8}{1-\frac{1}{2}}=16.$$

For a negative-ratio example, $8-4+2-1+\cdots$ has $r=-\frac{1}{2}$. Its partial sums oscillate above and below the limit, but the oscillations become smaller. Its sum is

$$S_\infty=\frac{8}{1+\frac{1}{2}}=\frac{16}{3}.$$

A common misconception is that convergence requires $0<r<1$. Negative ratios also converge when their magnitudes shrink. Another misconception is that terms approaching zero always prove convergence; this is not true for series in general.

IB Exam Technique

This is IB Mathematics AA SL 1.8 and may appear on Paper 1 or Paper 2. Before using the infinite-sum formula, explicitly verify $\lvert r\rvert<1$. State both the condition and the resulting sum; applying the formula without checking convergence can lose a method or reasoning mark.