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Math Coach Amy
Amy Ferguson Moncure

Calculus BC · Unit 10

Geometric Series

Add an infinite geometric series only when the ratio is strictly between −1 and 1. The first term in the sum is a.

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Key Ideas Easy Medium Hard

\(\sum_{n=0}^{\infty} ar^n=\dfrac{a}{1-r},\quad |r|<1\)

What to Remember

The infinite sum

If \(|r|<1\), then

\(\sum_{n=0}^{\infty} ar^n=\dfrac{a}{1-r}\).

\(a\) is the first term you add, and \(r\) is the common ratio.

A later starting index

If the sum starts at \(n=1\) or \(n=2\), use the first term that is actually in the sum. Do not automatically use the \(n=0\) term.

When it diverges

If \(|r|\ge 1\), the geometric series diverges. The terms do not shrink.

Repeating decimals

A repeating block is a geometric series. Write the first block over the right power of 10, then use \(r=1/100\) or \(r=1/10\) to match the block length.

Level 1

Easy

A ratio whose absolute value is clearly less than 1, or clearly not.

Example
Find the sum \(\sum_{n=0}^{\infty}\left(\dfrac{1}{2}\right)^n\).

\(S=\dfrac{a}{1-r}\)

\(a=1\) and \(r=\dfrac{1}{2}\)

\(S=\dfrac{1}{1-\dfrac{1}{2}}\)

\(S=\dfrac{1}{\dfrac{1}{2}}\)

\(S=2\)

Answer: \(2\)

1
Find \(\sum_{n=1}^{\infty}\left(\dfrac{1}{2}\right)^n\).
Show Answer

The first term is \(a=\dfrac{1}{2}\), and \(r=\dfrac{1}{2}\).

\(S=\dfrac{\frac{1}{2}}{1-\frac{1}{2}}\)

\(S=\dfrac{\frac{1}{2}}{\frac{1}{2}}\)

\(S=1\)

Answer: \(1\)

2
Find \(\sum_{n=0}^{\infty}\left(\dfrac{1}{3}\right)^n\).
Show Answer

\(S=\dfrac{1}{1-\dfrac{1}{3}}\)

\(S=\dfrac{1}{\dfrac{2}{3}}\)

\(S=\dfrac{3}{2}\)

Answer: \(\dfrac{3}{2}\)

3
Find \(\sum_{n=0}^{\infty} 4\left(\dfrac{1}{5}\right)^n\).
Show Answer

\(a=4\) and \(r=\dfrac{1}{5}\)

\(S=\dfrac{4}{1-\dfrac{1}{5}}\)

\(S=\dfrac{4}{\dfrac{4}{5}}\)

\(S=5\)

Answer: \(5\)

4
Does \(\sum_{n=0}^{\infty}\left(\dfrac{3}{2}\right)^n\) converge?
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\(r=\dfrac{3}{2}\)

\(|r|=\dfrac{3}{2}\ge 1\)

The series diverges.

Answer: diverges

5
Which series converges?
A. \(\sum \left(\dfrac{5}{4}\right)^n\)
B. \(\sum (-1)^n\)
C. \(\sum \left(\dfrac{2}{7}\right)^n\)
D. \(\sum 3^n\)
Show Answer

A has \(|r|=\dfrac{5}{4}>1\).

B has \(|r|=1\).

C has \(|r|=\dfrac{2}{7}<1\).

D has \(|r|=3>1\).

Only C converges.

Answer: C. \(\sum \left(\dfrac{2}{7}\right)^n\)

Level 2

Medium

A repeating decimal, a negative ratio, or a sum that does not start at n = 0.

Example
Write \(0.272727\ldots\) as a reduced fraction.

\(0.272727\ldots=\dfrac{27}{100}+\dfrac{27}{10000}+\dfrac{27}{1000000}+\cdots\)

\(a=\dfrac{27}{100}\) and \(r=\dfrac{1}{100}\)

\(S=\dfrac{\frac{27}{100}}{1-\frac{1}{100}}\)

\(S=\dfrac{\frac{27}{100}}{\frac{99}{100}}\)

\(S=\dfrac{27}{99}\)

\(S=\dfrac{3}{11}\)

Answer: \(\dfrac{3}{11}\)

1
Write \(0.363636\ldots\) as a reduced fraction.
Show Answer

\(a=\dfrac{36}{100}\) and \(r=\dfrac{1}{100}\)

\(S=\dfrac{\frac{36}{100}}{\frac{99}{100}}\)

\(S=\dfrac{36}{99}\)

\(S=\dfrac{4}{11}\)

Answer: \(\dfrac{4}{11}\)

2
Find \(\sum_{n=0}^{\infty} 5\left(-\dfrac{1}{4}\right)^n\).
Show Answer

\(a=5\) and \(r=-\dfrac{1}{4}\)

\(|r|=\dfrac{1}{4}<1\)

\(S=\dfrac{5}{1-\left(-\dfrac{1}{4}\right)}\)

\(S=\dfrac{5}{\dfrac{5}{4}}\)

\(S=4\)

Answer: \(4\)

3
Find \(\sum_{n=2}^{\infty}\left(\dfrac{1}{3}\right)^n\).
Show Answer

The first term in the sum is \(\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\).

\(r=\dfrac{1}{3}\)

\(S=\dfrac{\frac{1}{9}}{1-\frac{1}{3}}\)

\(S=\dfrac{\frac{1}{9}}{\frac{2}{3}}\)

\(S=\dfrac{1}{6}\)

Answer: \(\dfrac{1}{6}\)

4
A ball is dropped from 16 feet. Each bounce reaches half the height of the previous bounce. How far does the ball travel?
Show Answer

Downward distances: \(16+8+4+\cdots\)

\(S_{\text{down}}=\dfrac{16}{1-\frac{1}{2}}\)

\(S_{\text{down}}=32\)

Upward distances: \(8+4+2+\cdots\)

\(S_{\text{up}}=\dfrac{8}{1-\frac{1}{2}}\)

\(S_{\text{up}}=16\)

Total distance: \(32+16\)

\(48\) feet

Answer: \(48\) feet

5
Find the sum of the first four terms of \(\sum_{n=0}^{\infty}\left(\dfrac{1}{2}\right)^n\). Call that partial sum \(S_4\).
Show Answer

\(S_n=\dfrac{a(1-r^n)}{1-r}\)

\(a=1\), \(r=\dfrac{1}{2}\), and \(n=4\)

\(S_4=\dfrac{1-\left(\frac{1}{2}\right)^4}{1-\frac{1}{2}}\)

\(S_4=\dfrac{1-\frac{1}{16}}{\frac{1}{2}}\)

\(S_4=\dfrac{\frac{15}{16}}{\frac{1}{2}}\)

\(S_4=\dfrac{15}{8}\)

Answer: \(\dfrac{15}{8}\)

Level 3

Hard

A shifted index, a decimal with a new block, or the remainder of a geometric series.

Example
A prize pays 200 dollars at the end of the first year. Each later year it pays \(\dfrac{3}{4}\) of the previous year’s payment, forever. What is the total amount paid?

Payments: \(200+200\cdot\dfrac{3}{4}+200\cdot\left(\dfrac{3}{4}\right)^2+\cdots\)

\(a=200\) and \(r=\dfrac{3}{4}\)

\(S=\dfrac{200}{1-\frac{3}{4}}\)

\(S=\dfrac{200}{\frac{1}{4}}\)

\(S=800\)

The total is 800 dollars.

Answer: 800 dollars

1
Find \(\sum_{n=1}^{\infty} 3\left(\dfrac{2}{3}\right)^n\).
Show Answer

The \(n=1\) term is \(3\cdot\dfrac{2}{3}=2\).

\(r=\dfrac{2}{3}\)

\(S=\dfrac{2}{1-\frac{2}{3}}\)

\(S=\dfrac{2}{\frac{1}{3}}\)

\(S=6\)

Answer: \(6\)

2
Write \(0.454545\ldots\) as a reduced fraction.
Show Answer

\(a=\dfrac{45}{100}\) and \(r=\dfrac{1}{100}\)

\(S=\dfrac{\frac{45}{100}}{\frac{99}{100}}\)

\(S=\dfrac{45}{99}\)

\(S=\dfrac{5}{11}\)

Answer: \(\dfrac{5}{11}\)

3
For which real \(x\) does \(\sum_{n=0}^{\infty}\left(\dfrac{x}{4}\right)^n\) converge? Give the interval.
Show Answer

This is geometric with \(r=x/4\).

\(\left|\dfrac{x}{4}\right|<1\)

\(|x|<4\)

The interval is \((-4,4)\).

The endpoints \(x=4\) and \(x=-4\) make \(|r|=1\), so they diverge.

Answer: \((-4,4)\)

4
The infinite sum \(\sum_{n=0}^{\infty}\left(\dfrac{1}{2}\right)^n\) equals 2. After the four terms that make \(S_4=\dfrac{15}{8}\), what exact remainder is left?
Show Answer

The first omitted term is \(\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}\).

The tail is itself geometric with \(a=\dfrac{1}{16}\) and \(r=\dfrac{1}{2}\).

\(R=\dfrac{\frac{1}{16}}{1-\frac{1}{2}}\)

\(R=\dfrac{\frac{1}{16}}{\frac{1}{2}}\)

\(R=\dfrac{1}{8}\)

Check: \(2-\dfrac{15}{8}=\dfrac{1}{8}\).

Answer: \(\dfrac{1}{8}\)

5
Find \(\sum_{n=0}^{\infty} 6\left(-\dfrac{2}{5}\right)^n\).
Show Answer

\(a=6\) and \(r=-\dfrac{2}{5}\)

\(|r|=\dfrac{2}{5}<1\)

\(S=\dfrac{6}{1-\left(-\dfrac{2}{5}\right)}\)

\(S=\dfrac{6}{\dfrac{7}{5}}\)

\(S=\dfrac{30}{7}\)

Answer: \(\dfrac{30}{7}\)

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