Problem 55 - Algebra and Trigonometry

Question

Determine whether the infinite geometric series is convergent or divergent. If it is convergent, find its sum. $$ 1+\frac{3}{2}+\left(\frac{3}{2}\right)^{2}+\left(\frac{3}{2}\right)^{3}+\cdots $$

Step-by-Step Solution

Answer

The series is divergent.

Step 1: Identify the first term and common ratio

For the given series \(1 + \frac{3}{2} + \left( \frac{3}{2} \right)^2 + \cdots\), the first term \(a\) is 1. The common ratio \(r\) can be found by dividing the second term by the first term: \(\frac{3}{2}\).

Step 2: Determine if the series converges or diverges

A geometric series converges if the absolute value of the common ratio \(|r| < 1\). For this series, \(|r| = \left| \frac{3}{2} \right| = \frac{3}{2}\), which is greater than 1.

Step 3: Conclusion about the series

Since \(|r| = \frac{3}{2} > 1\), the series is divergent. Therefore, it does not have a sum.

Key Concepts

Infinite Series

An infinite series is a sum of infinite terms or numbers. When you see the term "infinite", it might sound overwhelming, but it's fascinating too! When terms keep adding up indefinitely, we call it an infinite series. This can either stretch towards infinity or approach a specific number. An infinite series can serve many purposes in mathematics and science:

• Represent functions or numbers compactly.
• Help solve equations with recurring patterns.

In our problem, we have an infinite geometric series: \[1 + \frac{3}{2} + \left(\frac{3}{2}\right)^{2} + \left(\frac{3}{2}\right)^{3} + \cdots\]Understanding the behavior of these series is key to determining if they possess finite sums.

Common Ratio

In a geometric series, the crucial component that links terms together is the common ratio. This is a consistent multiplier that helps generate each term from the previous one. For our series:

• First term, \(a\), is 1.
• To find the common ratio \(r\), divide any term by its previous term.
• Here, \(r = \frac{3}{2}\).

The common ratio plays a pivotal role in determining the nature of an infinite series. If \(|r| < 1\), the terms get smaller, leading towards convergence. Once \(|r|\) exceeds 1, each term increases, implying divergence. In our series, since \(r = \frac{3}{2}\), it exceeds 1, hinting towards a divergent direction.

Convergent Series

A convergent series has terms that approach zero as more are added. This series "settles down" to a particular value despite its infinite nature. For a geometric series:

• Convergence occurs if \(|r| < 1\).
• The series sum is given by the formula \( S = \frac{a}{1-r} \), where \(a\) is the first term.

Such series are meaningful because they can model real-world scenarios where values stabilize. Examples include decaying signals or financial depreciations over time.In our problem context, if \(r\) had been less than 1, we could've used this formula. However, since \(r = \frac{3}{2}\), the series cannot be convergent, and no sum exists.

Divergent Series

When a series diverges, it behaves erratically or grows without bound. This is the hallmark of a divergent series where the sum doesn't settle to a value.For geometries with \(|r| \ge 1\):

• Each successive term is as large as or larger than the one before.
• Sometimes these series reach infinity.

In practical terms, divergent series can be less useful, but they highlight growth scenarios. Imagine a situation where a population doubles each year without constraint — this is divergent behavior.Our series exhibits such divergence because \(r = \frac{3}{2}\), larger than 1. Hence, there's no meaningful sum — each term just keeps climbing. Understanding divergence adds depth to recognizing when growth is unsustainable in various fields.