Module 5.5 · Lesson 5.5.3

Sequences & Series — Infinite Sums & Convergence

Discover how adding infinitely many shrinking pieces can still settle on a single finite total, instead of growing forever.

Intuition

Walk halfway to a wall. Then walk half of what's left. Then half of what's left after that. You take infinitely many steps, yet you never overshoot the wall — and the total distance walked settles on one exact number. That's the heart of an infinite series: adding endlessly many terms that shrink fast enough to add up to something finite.

Core concept

A sequence is just an ordered list of numbers, one per index $n$. A series is what you get by adding the sequence's terms together, one after another. The running total after $n$ terms is called a partial sum, $S_n$. If those partial sums approach a specific number as $n$ grows, the series converges to that number; if the partial sums grow without bound instead, the series diverges.

½ + ¼ + ⅛ + ... → 1
Terms1/2, 1/4, 1/8, 1/16
Partial sum after 4 terms15/16 = 0.9375
Formula: a / (1 − r)(1/2) / (1 − 1/2)
Simplify(1/2) / (1/2)
AnswerSum converges to 1

Interactive Lab

Slide to add more terms of $\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots$ to the number line. Each new piece fills half of whatever gap remains — watch the total creep toward 1 without ever quite reaching it.

0 1
n = 4 partial sum Sₙ = 0.9375 gap remaining = 0.0625

Key Idea

$$ \sum_{n=1}^{\infty} a\,r^{\,n-1} = \frac{a}{1-r}, \qquad |r| < 1 $$

This is the geometric series formula. Every term is the previous term multiplied by the same ratio $r$. As long as that ratio's size is less than 1 — meaning each new piece is smaller than the last — the infinite sum settles on the finite value $a/(1-r)$ instead of growing forever.

$a$The first term of the series
$r$The common ratio between one term and the next
$|r| < 1$The condition required for the infinite sum to converge
$\sum_{n=1}^{\infty}$Add up every term, starting at $n=1$ and continuing forever

Think Further

  1. Predict what happens to the partial sums if the ratio were $r = 1.5$ instead of $r = 0.5$ — terms growing instead of shrinking. Would the series still converge?
  2. Compound interest sums up many small interest payments over a large number of periods. Why might treating that as an infinite series of shrinking contributions still give a very accurate approximation?
  3. Using the pattern from the worked example, predict the sum of $1 + \tfrac{1}{3} + \tfrac{1}{9} + \tfrac{1}{27} + \cdots$ without computing several partial sums by hand.
Show suggested answers
  1. No — with $r = 1.5$, each term is larger than the one before, so the partial sums keep growing without bound. The formula $a/(1-r)$ only applies when $|r| < 1$; outside that range the series diverges.
  2. As the number of compounding periods grows very large, each individual period's contribution to the total becomes tiny, behaving like the shrinking terms of a convergent series — so the infinite-sum formula closely tracks the true, very-large-but-finite total.
  3. Using $a/(1-r)$ with $a=1$ and $r=1/3$ gives $1 / (1 - 1/3) = 1 / (2/3) = 1.5$.