Problem 51 - Precalculus Mathematics for Calculus

Question

Determine whether the infinite geometric series is convergent or divergent. If it is convergent, find its sum. $$1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots$$

Step-by-Step Solution

Answer

The series is convergent and its sum is \(\frac{3}{2}\).

Step 1: Identify the First Term and Common Ratio

The series given is \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\). The first term \(a\) is \(1\). Each subsequent term is obtained by multiplying the previous term by \(\frac{1}{3}\). Thus, the common ratio \(r\) is \(\frac{1}{3}\).

Step 2: Determine Convergence Using the Common Ratio

A geometric series converges if the absolute value of the common ratio \(|r|\) is less than 1. Here, \(|r| = \left| \frac{1}{3} \right| = \frac{1}{3} < 1\). Therefore, the series is convergent.

Step 3: Use the Formula for the Sum of an Infinite Geometric Series

The sum \(S\) of an infinite geometric series with first term \(a\) and common ratio \(r\) (where \(|r| < 1\)) is given by the formula: \[ S = \frac{a}{1 - r} \]. Substitute \(a = 1\) and \(r = \frac{1}{3}\) into the formula to get: \[ S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \].

Key Concepts

Infinite Series

An infinite series is a sum of infinitely many numbers. In the context of a geometric series, it is the sum of numbers where each term is a constant multiple of the previous one. This can look like

• 1 + \( r \) + \( r^2 \) + \( r^3 \) + \( \ldots \)

In our exercise, the series is \( 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \). Here, the common multiple is the same "r" for each term—in our case, that is \( \frac{1}{3} \).
An infinite series is a fascinating concept because, unlike regular sums that end, infinite series extend infinitely but can still be manageable. Understanding them is crucial in core mathematical concepts.

Convergence

Convergence is a critical aspect of studying series. For a series to be convergent, the sum of its terms must approach a finite number as you add more and more terms. With geometric series, whether it converges depends on the absolute value of the common ratio \( r \).

• If \( |r| < 1 \), the series converges, meaning its sum approaches a fixed value.
• If \( |r| \geq 1 \), the series diverges, and its sum increases without bound.

In our example, with a common ratio of \( \frac{1}{3} \), the series converges because \( \frac{1}{3} < 1 \). This means, as you keep adding terms like \( \frac{1}{9} \) and \( \frac{1}{27} \), you approach a specific sum rather than heading towards infinity.

Sum of a Series

Finding the sum of a convergent geometric series helps make infinite series more tangible. For such series, the sum \( S \) can be found using the formula: \[ S = \frac{a}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio.
Let's look at our series \( 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \). Here, the first term \( a \) is \( 1 \) and \( r = \frac{1}{3} \). When we plug these into the formula, we get: \[ S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \].
This calculation tells us that the terms of this series add up to \( \frac{3}{2} \) as you keep summing them indefinitely. This clear sum exemplifies the power of understanding convergence and applying the sum formula appropriately.