Problem 19 - Calculus: One and Several Variables

Question

Show that the series diverges. $$1+\frac{3}{2}+\frac{9}{4}+\frac{27}{8}+\frac{81}{16}+\dots$$

Step-by-Step Solution

Answer

The given series has the general term \(a_n = \frac{3^n}{2^n}\), making it a geometric series with a common ratio \(r = \frac{3}{2}\). Since |r| > 1, the series diverges by the geometric series convergence criteria.

Step 1: Identify the series as a geometric series

The given series can be written as: $$\sum_{n=0}^\infty a_n$$ where we need to find the general term \(a_n\) and the common ratio.

Step 2: Find the general term and common ratio

Take note that each term is a product of its numerator and denominator, which are both powers of 3 and 2, respectively. Notice that for the term \(n\), the numerator has \(3^n\) and the denominator has \(2^n\). So, the general term of the series is: $$a_n = \frac{3^n}{2^n}$$ The common ratio, r, of a geometric series is the ratio of a term to its preceding term. We can find r from the general term: $$r = \frac{a_{n+1}}{a_n} = \frac{\frac{3^{n+1}}{2^{n+1}}}{\frac{3^n}{2^n}} = \frac{3}{2}$$ r is equal to 3/2 in this case.

Step 3: Apply the geometric series convergence criteria

A geometric series converges if the absolute value of its common ratio is less than 1, i.e., |r| < 1. Otherwise, it diverges. In our case, the common ratio is |3/2| = 3/2, which is not less than 1.

Step 4: Conclude the result

Since the common ratio r = 3/2 and |r| > 1, we can conclude that the given series: $$1+\frac{3}{2}+\frac{9}{4}+\frac{27}{8}+\frac{81}{16}+\dots$$ diverges as per the geometric series convergence criteria.

Key Concepts

Divergence

Divergence is a key concept in mathematical series. It describes a series that does not have a finite sum. When a series doesn't settle to a specific value as it progresses, it's said to diverge.
In a divergent series, the terms might become larger as they are added, or they might not stabilize around any particular sum. Understanding divergence helps in determining whether specific series can be useful for calculations or models.

In the context of the series given, it diverges because its terms grow without bound, and thus do not converge to a finite sum. Identifying divergence early in analyzing a series saves us from misapplying formulas or methods suited for convergent series.

Series Convergence

Series convergence occurs when the terms of a series cumulatively approach a specific limit. In simple terms, a convergent series will balance out toward a certain sum as more terms are added.
This is crucial because only convergent series have a meaningful and finite sum, making them valuable for analysis and applications. The convergence of a series largely depends on its general term and the properties of the series.

Geometric series, like the one in the exercise, have specific criteria for convergence. For them, we look at the absolute value of the common ratio. When this value is less than 1, the series converges; when it's greater or equal to 1, it doesn't. In our problem, since the common ratio is greater than 1, the series fails the convergence test, confirming it's divergent.

Common Ratio

The common ratio in a geometric series is the factor between consecutive terms. It's a crucial element in determining the behavior of the series.
By understanding the common ratio, we can spot patterns and predict the future terms of the series.

In our geometric series, we calculate the common ratio by dividing any term by its previous one. For example, using the series given, the ratio is calculated as \( r = \frac{3}{2} \).
Since this ratio is consistent throughout the series, it governs the growth or shrinkage of the series.

For convergence analysis, the common ratio's absolute value must be less than 1 for the series to converge. Here, since \( |r| = \frac{3}{2} > 1 \), the common ratio indicates that our series will not converge, but rather diverge.