Infinite Series, Partial Sums, and the Geometric Series

Look back at the bars above with r = 0.5: each new term is half the previous one, so each bright cap is half as tall as the last. The staircase climbs 1, 1.5, 1.75, 1.875, … — always rising, but by shrinking steps, flattening into the ceiling at \(\frac{a}{1-r}\). Now drag r past 1: every cap is now taller than the one before, and the totals sprint off the chart. Convergence of a series is a picture about the bars (the running totals), not the individual terms.

Given a sequence \(\{a_n\}\) — the objects from the previous lesson — define the sequence of partial sums \(S_N = \sum_{n=1}^{N} a_n\). Each \(S_N\) is an ordinary finite sum, a single number. The series converges to \(S\) if \(\lim_{N\to\infty} S_N = S\) in the usual sequence sense; if \(\{S_N\}\) diverges, the series diverges. This reduces every series question to a sequence question about running totals — which is exactly why sequences are studied first. For the geometric series \(\sum_{n=0}^{\infty} ar^n\) with first term \(a \neq 0\) and common ratio \(r\), the partial sum has a closed form: \(S_N = a\,\frac{1-r^{N+1}}{1-r}\) for \(r \neq 1\). Everything hinges on \(r^{N+1}\), which tends to 0 exactly when \(|r| < 1\) — giving \(\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\). For \(|r| \ge 1\) the series diverges (at \(r = 1\) the partial sums grow like \(Na\); otherwise \(r^{N+1}\) never settles).

Why does one series family get its own lesson? Because the geometric series is essentially the only family with a simple, general closed-form sum. The convergence tests coming up — integral, comparison, ratio, root — only tell you whether a series converges, not what it converges to, and several of them work by comparing an unknown series against a geometric benchmark. Master \(\frac{a}{1-r}\) now and you hold the yardstick the rest of the chapter measures against.