The sum of the series \(3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots \) is equal to
9/4
Finding the Sum of an Infinite Geometric Series
The given series is \(3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots \). This is an infinite series where each term is obtained by multiplying the previous term by a constant value. This type of series is known as a geometric series.
Identifying the First Term and Common Ratio
In a geometric series, the first term is denoted by \(a\), and the constant multiplier is called the common ratio, denoted by \(r\).
- The first term is the initial term of the series: \(a = 3\).
- The common ratio \(r\) can be found by dividing any term by its preceding term. Let's calculate it using the first two terms:
\(r = \frac{\text{second term}}{\text{first term}} = \frac{-1}{3}\)
Let's verify this with the next pair of terms:
\(r = \frac{\text{third term}}{\text{second term}} = \frac{\frac{1}{3}}{-1} = -\frac{1}{3}\)
The common ratio is indeed \(r = -\frac{1}{3}\).
Condition for Convergence of an Infinite Geometric Series
An infinite geometric series converges to a finite sum if and only if the absolute value of the common ratio \(r\) is less than 1, i.e., \(|r| < 1\).
In this case, \(|r| = \left|-\frac{1}{3}\right| = \frac{1}{3}\).
Since \(\frac{1}{3} < 1\), the given series converges.
Formula for the Sum of a Convergent Infinite Geometric Series
The sum \(S\) of a convergent infinite geometric series is given by the formula:
\(S = \frac{a}{1-r}\)
where \(a\) is the first term and \(r\) is the common ratio.
Calculating the Sum of the Series
Now, we substitute the values of \(a\) and \(r\) into the formula:
\(a = 3\)
\(r = -\frac{1}{3}\)
\(S = \frac{3}{1 - \left(-\frac{1}{3}\right)}\)
\(S = \frac{3}{1 + \frac{1}{3}}\)
To simplify the denominator, we find a common denominator:
\(1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{3+1}{3} = \frac{4}{3}\)
So, the sum is:
\(S = \frac{3}{\frac{4}{3}}\)
Dividing by a fraction is equivalent to multiplying by its reciprocal:
\(S = 3 \times \frac{3}{4}\)
\(S = \frac{3 \times 3}{4}\)
\(S = \frac{9}{4}\)
Therefore, the sum of the given infinite geometric series \(3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots \) is \(\frac{9}{4}\).
| First Term (\(a\)) | \(3\) |
|---|---|
| Common Ratio (\(r\)) | \(-\frac{1}{3}\) |
| Condition for Convergence (\(|r| < 1\)) | \(\left|-\frac{1}{3}\right| = \frac{1}{3} < 1\) (Converges) |
| Sum Formula (\(S\)) | \(\frac{a}{1-r}\) |
| Calculated Sum | \(\frac{9}{4}\) |
Revision Table: Key Geometric Series Concepts
| Concept | Description | Formula/Condition |
|---|---|---|
| Geometric Series | A series where the ratio between consecutive terms is constant. | \(a, ar, ar^2, ar^3, \ldots\) |
| First Term (\(a\)) | The initial term of the series. | - |
| Common Ratio (\(r\)) | The constant factor between consecutive terms. | \(r = \frac{ar^n}{ar^{n-1}}\) |
| Infinite Geometric Series | A geometric series with an infinite number of terms. | \(a + ar + ar^2 + \ldots\) |
| Convergence Condition | An infinite geometric series converges if its common ratio's absolute value is less than 1. | \(|r| < 1\) |
| Sum of Convergent Infinite Series | The finite value the series approaches when it converges. | \(S = \frac{a}{1-r}\) |
Additional Information: Divergent Geometric Series
If the absolute value of the common ratio \(|r|\) for an infinite geometric series is greater than or equal to 1 (\(|r| \ge 1\)), the series diverges. This means the sum of the terms does not approach a finite value as more terms are added. Instead, the partial sums either grow infinitely large, infinitely small, or oscillate without approaching a single value.
For example, the series \(1 + 2 + 4 + 8 + \ldots\) has \(a=1\) and \(r=2\). Since \(|2| \ge 1\), the series diverges. The partial sums (1, 3, 7, 15, ...) grow without bound.
Another example is the series \(1 - 2 + 4 - 8 + \ldots\) which has \(a=1\) and \(r=-2\). Since \(|-2| \ge 1\), this series also diverges. The partial sums (1, -1, 3, -5, ...) oscillate and do not approach a single value.
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