Geometric series: nth term, partial sums, infinite sums
Geometric Series
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 derive and use the formula for the sum of a geometric series.
A geometric series sums a sequence with a constant ratio, with \(S_n=a_1\dfrac{1-r^n}{1-r}\) and, when \(|r|<1\), the infinite sum \(\dfrac{a_1}{1-r}\).
Theory
A geometric sequence has a constant common ratio \(r\): each term is the previous one times \(r\). Its \(n\)th term is
The sum of the first \(n\) terms is
and if \(|r|<1\) the terms shrink to zero and the infinite series converges:
Term, partial sum, and infinite sum:
How to work with geometric series
- Find \(r\) as the ratio of consecutive terms.
- nth term: \(a_1 r^{\,n-1}\).
- Partial sum: \(a_1\dfrac{1-r^n}{1-r}\).
- Infinite sum: \(\dfrac{a_1}{1-r}\) — only if \(|r|<1\).
\(a_1=3,\ r=2\); use \(a_n=a_1r^{n-1}\).
| a_6 | \(=\) | 3\cdot 2^{5} |
| \(=\) | 3\cdot 32=96 |
Use \(S_n=a_1\dfrac{1-r^n}{1-r}\).
| S_6 | \(=\) | 3\cdot\dfrac{1-2^6}{1-2} |
| \(=\) | 3\cdot\dfrac{-63}{-1}=189 |
\(a_1=4,\ r=\dfrac12\); since \(|r|<1\), use \(S_\infty=\dfrac{a_1}{1-r}\).
| S_\infty | \(=\) | \dfrac{4}{1-\dfrac12} |
| \(=\) | \dfrac{4}{\dfrac12}=8 |
Here \(r=3\) and \(|r|\ge 1\), so the terms grow — the series diverges.
| |r|=3\ge 1 | \(\Rightarrow\) | \text{diverges} |
Common pitfalls
Frequently asked questions
What is a geometric sequence?
A sequence with a constant ratio \(r\) between consecutive terms; the \(n\)th term is \(a_1 r^{\,n-1}\).
How do you sum a finite geometric series?
\(S_n=a_1\dfrac{1-r^n}{1-r}\) for \(r\neq 1\).
When does an infinite geometric series converge?
Only when \(|r|<1\); then \(S_\infty=\dfrac{a_1}{1-r}\). If \(|r|\ge 1\) it diverges.
What is the common ratio?
The fixed multiplier \(r\) taking each term to the next.