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Geometric Series
Calculus 2 · Chapter 6: Sequences & Introduction to SeriesSection 6.3 · lesson 29 of 5411 min · free
← Infinite Series and Partial SumsNext lesson: Telescoping Series →
In this lesson
The one family whose exact sum we can write down.
- Derive the partial-sum formula by the shift-and-subtract trick
- Apply the convergence condition |r|<1 and the sum a/(1-r)
- Identify a as the first term that actually appears
- Use it for repeating decimals, bouncing balls and drug levels
Transcript
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MAT-265 - Calculus 2 · Section 6.3
Geometric Series
Almost no infinite series has a sum you can write down exactly. The geometric series is the great exception. And that makes it the most important series in the course. A geometric series is one in which each term is a fixed multiple of the previous one. Sum of a, r to the n. Where r is called the common ratio. Because its convergence behavior is completely understood, and its sum is known exactly, it becomes the reference series against which other, more complicated series are compared. Several major convergence tests in later sections work by comparing an unfamiliar series to a geometric one. And power series — the infinite polynomials that represent functions like e to the x and sine x — are built directly on geometric reasoning. So mastering this now pays dividends well beyond this section.
The Derivation
Shift and Subtract
Let's find the partial sums, using a trick worth knowing. Write S sub N as a, plus a r, plus a r squared, and so on up to a r to the N minus one. Now multiply the whole thing by r. That shifts every term one place to the right. r S sub N is a r, plus a r squared, all the way to a r to the N. Now subtract the second from the first. And watch what happens. Almost everything cancels — every middle term appears in both lines. Only the first term of one and the last term of the other survive. S sub N minus r S sub N equals a, minus a r to the N. Factor the left side: S sub N times one minus r. So, provided r is not one, divide to get the partial sum formula. S sub N equals a times one minus r to the N, all over one minus r. That shift-and-subtract move is worth remembering. It's the standard way to collapse a geometric sum.
Taking the Limit
Everything Hinges on One Term
The infinite sum is the limit of S sub N as N goes to infinity. Now look carefully at that formula. The only part that depends on N is the term r to the N. So everything hinges on how r to the N behaves. And there are exactly two cases. Case one. If the absolute value of r is less than one, then r to the N goes to zero. Repeatedly multiplying by a number smaller than one in magnitude drives the product to zero. So the partial sums approach a finite limit, and the series converges. Case two. If the absolute value of r is at least one, then r to the N does not approach zero. It grows without bound when r exceeds one in size. It fails to settle when r is minus one. And it stays constant when r is one. In every one of those cases the partial sums do not converge, and the series diverges. So substituting r to the N goes to zero into the partial sum formula gives the key result. The geometric series converges if and only if the absolute value of r is less than one. And in that case its sum is a, over one minus r. That formula deserves emphasis. The sum depends only on the first term and the common ratio. The numerator is the term the series starts from. The denominator measures how quickly the terms shrink.
Worked Examples
Identify a and r, Then Check |r|
Three examples, and each teaches a different lesson. First. Evaluate the sum of eight times one half to the n, starting at n equals zero. That's eight plus four plus two plus one plus one half, and so on. Identify the pieces. The first term a is eight. The common ratio r is one half. Since the absolute value of r is less than one, the series converges. Apply the formula: eight over one minus one half. Which is eight over one half. Sixteen. Infinitely many positive terms add up to exactly sixteen. Second. Does the sum of three times two to the n converge? Here a is three and r is two. Since the absolute value of r is two, which is at least one, the series diverges. The terms grow without bound. And the sum formula does not apply. Writing three over one minus two, which is minus three, would be meaningless — a sum of positive terms cannot be negative. So checking the absolute value of r before applying the formula prevents that error. And a negative answer for a positive series is the tell-tale sign you skipped the check. Third, and this is the one that catches people. Evaluate the sum of five times one third to the n — starting at n equals one, not zero. So the first term is not five. It's five times one third. Which is five thirds. The common ratio is still one third. Use the first actual term as a. Five thirds over one minus one third. Which is five thirds over two thirds. Five halves. The key is that a in the formula always means the first term that actually appears. Whatever the starting index.
Application 1
Every Repeating Decimal Is a Geometric Series
The exact-sum formula makes geometric series directly useful anywhere a quantity is repeatedly scaled by a fixed fraction. Start with a fact you may never have questioned. Every repeating decimal equals a fraction. Why? Because every repeating decimal is secretly a geometric series. The repeating block, shifted further right each time, forms terms in a fixed ratio. A power of one tenth for each digit in the block. Let's express zero point seven repeating as a fraction. Write the decimal as a sum of its pieces. Seven tenths, plus seven hundredths, plus seven thousandths, and so on. That's geometric, with a equal to seven tenths and r equal to one tenth. Apply the formula. Seven tenths over one minus one tenth. Which is seven tenths over nine tenths. Seven ninths. And a quick check confirms it. Seven divided by nine is zero point seven seven seven. Now a two-digit block. Express zero point four five repeating. The block forty-five occupies two decimal places, so each copy is scaled by one hundredth. So a is forty-five hundredths, and r is one hundredth. Apply the formula and you get forty-five over ninety-nine. Which simplifies to five elevenths. The pattern generalizes. One tenth for a one-digit block. One hundredth for two digits. One thousandth for three.
Applications 2 and 3
A Bouncing Ball, and a Steady Drug Level
Two more applications, both physical. A ball is dropped from a height of ten feet. On each bounce it returns to three quarters of its previous height. Find the total vertical distance it travels before coming to rest. Careful accounting matters here. The ball first falls ten feet. That's one trip. Then each bounce contributes two trips — up to a height, and back down. The bounce heights are ten times three quarters, then ten times three quarters squared, and so on. Each traveled twice. So the total distance is the initial drop, plus twice the sum of the bounce heights. That series has first actual term ten times three quarters, which is seven point five. And ratio three quarters. Its sum is seven point five over one quarter. Thirty. So the total distance is ten plus two times thirty. Seventy feet. Despite bouncing infinitely many times, the ball travels a finite seventy feet — because the heights form a convergent geometric series. Last application. Steady-state drug levels. A patient takes a hundred milligram dose once daily. Each day the body eliminates forty percent of the drug present, leaving sixty percent. Find the steady-state amount in the body just after a dose, in the long run. Just after the first dose there's a hundred milligrams. By the next dose, sixty percent remains — that's a hundred times zero point six. Then a new hundred milligrams is added. Tracing the amount present right after each dose, the accumulated residue from all previous doses forms a geometric series with a equal to one hundred and r equal to zero point six. Since the ratio is less than one, the long-run level converges. One hundred over zero point four. Two hundred fifty milligrams. So the drug level stabilizes at two hundred fifty milligrams just after each dose. It does not build up indefinitely — because fixed-fraction clearance exactly balances the daily intake in the limit. That steady-state reasoning is central to how dosing schedules are designed.
Key Takeaways
What to Carry Forward
Here's something worth thinking about. The convergence condition, absolute r less than one, looks structurally similar to a result from Chapter Two. The integral from one to infinity of x to the minus p converges exactly when p is greater than one. Both describe a sharp threshold. Separating the pieces shrink fast enough to sum to something finite, from they shrink too slowly. For the geometric series, why is the boundary case, absolute r equals one, excluded? What specifically happens to the terms a r to the n when r is one, or when r is minus one? And why does that prevent the partial sums from ever settling on a limit? Five takeaways. A geometric series has a constant ratio r between consecutive terms. Its partial sum comes from shifting and subtracting. It converges if and only if the absolute value of r is less than one, with sum a over one minus r. One of the very few series whose exact sum is known in closed form. a is the first term that actually appears, whatever the starting index. A series beginning at n equals one has a equal to a r, not a. Always check the absolute value of r before applying the formula. A negative sum for a positive series means the check was skipped. And repeating decimals, bouncing balls, and steady-state drug levels are all geometric series in disguise. Next up is section six point four — Telescoping Series. The other family whose partial sums collapse into something you can actually evaluate.
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