Problem 42 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\left(2-\frac{1}{2^{n}}\right)\left(3+\frac{1}{2^{n}}\right) $$

Step-by-Step Solution

Answer

The sequence converges to 6.

Step 1: Expand the Product

To analyze the sequence, we first expand the expression \(a_n = \left(2 - \frac{1}{2^n}\right)\left(3 + \frac{1}{2^n}\right)\) using the distributive property. Compute: \[\left(2 - \frac{1}{2^n}\right)\left(3 + \frac{1}{2^n}\right) = 2 \times 3 + 2 \times \frac{1}{2^n} - \frac{1}{2^n} \times 3 - \frac{1}{2^n} \times \frac{1}{2^n}\]Simplifying this, we get:\[6 + \frac{2}{2^n} - \frac{3}{2^n} - \frac{1}{4^n}\]

Step 2: Simplify the Expression

We simplify the expression:\[6 + \frac{2}{2^n} - \frac{3}{2^n} - \frac{1}{4^n} = 6 - \frac{1}{2^n} - \frac{1}{4^n}\]So, the sequence \(a_n\) becomes:\[a_n = 6 - \frac{1}{2^n} - \frac{1}{4^n}\]

Step 3: Determine the Limit as \(n\to\infty\)

To find whether the sequence converges, we examine the limit of \(a_n\) as \(n\) approaches infinity. Consider each term:- As \(n \to \infty\), \(\frac{1}{2^n} \to 0\).- As \(n \to \infty\), \(\frac{1}{4^n} \to 0\).Thus, the limit of \(a_n\) is:\[\lim_{{n \to \infty}} a_n = 6 - 0 - 0 = 6\]

Step 4: Conclusion on Convergence

Since the limit \(\lim_{{n \to \infty}} a_n = 6\) exists and is finite, the sequence \(\{a_n\}\) converges to 6.

Key Concepts

Limit of a Sequence

When discussing the limit of a sequence, we are essentially trying to determine what value a sequence approaches as its index tends to infinity. For the sequence \(a_n = 6 - \frac{1}{2^n} - \frac{1}{4^n}\), we evaluate each component separately. These tiny fractions \left(\frac{1}{2^n} \text{ and } \frac{1}{4^n}\right)\ shrink as \(n\) gets larger.

So as \(n o \infty\):

• \(\frac{1}{2^n}\) approaches 0
• \(\frac{1}{4^n}\) approaches 0

Thus, \(a_n\) converges to \(6 - 0 - 0 = 6\). This shows that the limit of the sequence \(\left\{a_n\right\}\) is \(6\). Limits give us a clear understanding of the sequence's behavior as it progresses towards infinity.

Distributive Property

The distributive property is a fundamental principle in algebra, allowing the expansion of expressions, like \( (a-b)(c+d) = ac + ad - bc - bd\). By applying this to \(a_n = (2 - \frac{1}{2^n})(3 + \frac{1}{2^n})\), we expand it step by step:

• \(2 imes 3 = 6\)
• \(2 imes \frac{1}{2^n} = \frac{2}{2^n}\)
• \(-\frac{1}{2^n} imes 3 = -\frac{3}{2^n}\)
• \(-\frac{1}{2^n} imes \frac{1}{2^n} = -\frac{1}{4^n}\)

By using the distributive property, we transform the complex form into a simpler and more manageable sequence expression: \(6 + \frac{2}{2^n} - \frac{3}{2^n} - \frac{1}{4^n}\), which further simplifies to \(6 - \frac{1}{2^n} - \frac{1}{4^n}\).

This simplification makes it easier to analyze the sequence's convergence and find the limit.

Infinite Limits

'Infinite limits' can sound intimidating, but they're an essential part of understanding sequences. They occur when a sequence's value grows without bounds (positive or negative) as the index increases. However, with sequences where terms shrink indefinitely, like with \(\frac{1}{2^n}\) or \(\frac{1}{4^n}\), they tend to zero.

In this context, reaching zero means that as \(n\) becomes very large, these fractional components disappear from the expression effectively.

• It ensures the sequence doesn't grow to infinity in either direction.
• Instead, it stabilizes around a finite number, confirming convergence.

Thus, understanding infinite limits provides insight into the long-term behavior of sequences.

Convergent Sequences

A sequence is convergent when it approaches a single, finite value as \(n o \infty\). In the case of \(\{a_n\} = 6 - \frac{1}{2^n} - \frac{1}{4^n}\), it converges to 6. Why? Because the decreasing terms \(\frac{1}{2^n} \text{ and } \frac{1}{4^n}\) become negligible.

Convergent sequences:

• Have a limit that exists (such as the number 6 in our example).
• Do not oscillate or move towards infinity.
• Simplify analysis by providing straightforward behavior as \(n\) grows large.

The finite limit \(6\) signifies that the sequence stabilizes over time, enabling predictions and further mathematical analysis. Recognizing convergent sequences is crucial in calculus and analysis, as they ensure predictable outcomes in various scenarios.