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Algebra 2 Sequences and series

Infinite geometric series

20 practice questions 0 video lessons Theory + worked examples

Infinite Geometric Series

Common Core Algebra 2 • Standard A-SSE.4 • Sequences & Series

Infinite Geometric Series is a topic in Sequences & Series in the Common Core State Standards. It is aligned to Standard A-SSE.4, which requires students to use the structure of a geometric series to find the sum, including the limiting sum when the ratio is less than one in size.

An infinite geometric series converges to \(\dfrac{a_1}{1-r}\) when \(|r|<1\); otherwise it diverges.

Common Core Algebra 2 › Sequences & Series › Infinite Geometric Series  —  Standard A-SSE.4

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Theory

An infinite geometric series adds infinitely many terms. It converges only when the ratio is small:

\[S=\dfrac{a_1}{1-r}\quad\text{if}\quad |r|<1.\]

If \(|r|\ge1\), the terms don't shrink and the series diverges.

Only shrinking terms (\(|r|<1\)) add to a finite total.
Partial sums converge When the ratio is between -1 and 1, the partial sums approach a finite limit. n sum → 16
Partial sums approach the limit \(16\).
Infinite geometric series Infinite geometric series Infinite geometric series converges only if |r| < 1 sum: S = a₁ / (1 - r) |r| ≥ 1: the series diverges
When an infinite series converges.

Sum to infinity:

\[S=\dfrac{a_1}{1-r},\qquad |r|<1\]
the infinite sum is the first term over one minus the ratio, when the ratio is less than one in size
Check \(|r|<1\) first — otherwise there is no sum.

How to sum an infinite series

  1. Find \(a_1\) and \(r\).
  2. Check that \(|r|<1\).
  3. Apply \(S=\dfrac{a_1}{1-r}\).
  4. If \(|r|\ge1\), state that it diverges.
Example 1 — A convergent sum
Find \(8+4+2+1+\dots\).
Solution

Here \(a_1=8,\ r=\dfrac12\), and \(|r|<1\).

\(S\)\(=\)\(\dfrac{8}{1-\dfrac12}\)
\(=\)\(\dfrac{8}{\dfrac12}=16\)
the sum is 16
Example 2 — Repeating decimal
Express \(0.\overline{9}\) as a fraction using a series.
Solution

\(0.9+0.09+\dots\) has \(a_1=0.9,\ r=0.1\).

\(S\)\(=\)\(\dfrac{0.9}{1-0.1}\)
\(=\)\(\dfrac{0.9}{0.9}=1\)
the sum is 1, so point nine repeating equals 1
Example 3 — Does it converge?
Does \(2+4+8+16+\dots\) have a finite sum?
Solution

The ratio \(r=2\) has \(|r|\ge1\).

\(|r|\)\(=\)\(2\ge1\)

It diverges — no finite sum.

no, it diverges because the ratio is at least 1
Example 4 — Find the ratio limit
Find \(1+\dfrac13+\dfrac19+\dots\).
Solution

\(a_1=1,\ r=\dfrac13\).

\(S\)\(=\)\(\dfrac{1}{1-\dfrac13}\)
\(=\)\(\dfrac{1}{\dfrac23}=\dfrac32\)
the sum is three halves

Common pitfalls

Converges only if \(|r|<1\) — always check first.
Use \(a_1\), the first term, in the formula.
\(|r|\ge1\) means no finite sum.

Frequently asked questions

When does an infinite geometric series converge?

When \(|r|<1\).

What is the sum to infinity?

\(S=\dfrac{a_1}{1-r}\).

What happens if \(|r|\ge1\)?

The series diverges — there is no finite sum.

Why does \(0.\overline9=1\)?

The geometric series \(0.9+0.09+\dots\) sums to exactly \(1\).

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