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

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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 $\lvert r\rvert<1$.

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,

For example,

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

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

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.