Problem 46 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\frac{1}{(0.9)^{n}} $$

Step-by-Step Solution

Answer

The sequence diverges and does not have a limit.

Step 1: Identify the Sequence

The sequence given is \( a_n = \frac{1}{(0.9)^n} \). It is a geometric sequence because each term is divided by \( 0.9 \) to get the next term.

Step 2: Understand the Behavior of the Sequence

A geometric sequence \( a_n = ar^n \) converges if the absolute value of the common ratio \( r < 1 \). Here, the common ratio is \( r = 0.9 \), which is less than 1. However, in our sequence, it's \( \frac{1}{(0.9)^n} = (0.9)^{-n} \), and cubbing by 0.9 then to a negative exponent will give a diverging sequence (actually, \( r = \frac{1}{0.9} > 1 \)). Thus, this sequence diverges.

Step 3: Determine Sequence Behavior

Since the base \((0.9)^n\) is getting progressively smaller, its reciprocal, \( \frac{1}{(0.9)^n} \), gets larger and larger. As \( n \to \infty \), \( (0.9)^n \to 0 \), so \( \frac{1}{(0.9)^n} \to \infty \). Thus, the sequence \( a_n \) diverges.

Key Concepts

Sequence Convergence

In mathematics, not all sequences approach a finite value as they progress. However, when they do, we say they "converge." Convergence is a fundamental concept when studying sequences. For a sequence to converge, its terms must approach a single finite number as the number of terms goes to infinity.

For a more technical approach, a sequence \( \{a_n\} \) is said to converge to a limit \( L \) if, for any chosen small positive number \( \varepsilon \), there exists a point beyond which all terms of the sequence fall within \( \varepsilon \) of \( L \). This is written in formal terms as:

• \( \lim_{{n \to \infty}} a_n = L \)

In the original problem, we evaluated the sequence \( a_n = \frac{1}{(0.9)^n} \). To check for convergence, it's crucial to examine the behavior of the sequence as \( n \) grows very large. Here, because the base \( (0.9)^n \) becomes exceedingly small, \( \frac{1}{(0.9)^n} \) rapidly grows, indicating divergence instead.

Divergence of Sequences

When discussing the divergence of a sequence, we refer to sequences that do not settle on a fixed limit. They either increase or decrease without bound, or they oscillate instead of converging to a single value.

In exploring the original sequence \( \left\{ a_{n} \right\} = \frac{1}{(0.9)^n} \), we discover it's actually a geometric sequence where the base exponent falls under the inverse effect of \( 0.9 \). Geometrically, the sequence is expressed as \((0.9)^{-n}\). As \( n \) gets larger, \( (0.9)^n \) approaches zero.

• This makes \( \frac{1}{(0.9)^n} \) grow without limit, effectively approaching infinity.
• Since this sequence does not approach a particular finite number, we describe it as divergent.

Limits of Sequences

Finding the limit of a sequence is pivotal when determining whether a sequence converges or diverges. For sequences that converge, the limit gives us the final value the sequence approaches.

The general method for finding a limit involves analyzing the pattern or formula of the sequence. If by letting \( n \to \infty \), the terms of the sequence approach a specific number, that number is the sequence's limit.

In cases where the sequence diverges, like our given sequence \( \left\{ a_n \right\} = \frac{1}{(0.9)^n} \), the sequence does not settle towards any fixed value. Therefore, it doesn't have a limit in the typical sense. Instead, we describe the behavior as approaching infinity since the sequence's terms continue to grow larger.