Problem 9

Question

In the following exercises, feel free to use what you know from calculus to find the limit, if it exists. But you must prove that you found the correct limit, or prove that the series is divergent. Show that the sequence $\left\\{\frac{1}{\sqrt[3]{n}}\right\\}$ is monotone, bounded, and use Proposition 2.1.10 to find the limit.

Step-by-Step Solution

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Answer

The sequence is decreasing, bounded, and converges to 0.

1Step 1: Prove Monotonicity

To show the sequence $ a_n = \frac{1}{\sqrt[3]{n}} $ is monotone, we check if it is decreasing. Compare $ a_{n+1} $ and $ a_n $:\[\frac{1}{\sqrt[3]{n+1}} \leq \frac{1}{\sqrt[3]{n}}.\] By rearranging, this reduces to $ n \leq n+1 $, which is always true. Hence, the sequence is monotonically decreasing.

2Step 2: Prove Boundedness

To show $ a_n = \frac{1}{\sqrt[3]{n}} $ is bounded, observe $ a_n $ is positive since $ \sqrt[3]{n} > 0 $ for all $ n > 0 $. We need an upper bound: $ a_1 = \frac{1}{\sqrt[3]{1}} = 1 $ is an upper bound because each subsequent term is less than or equal to 1. Thus, the sequence is bounded below by 0 and above by 1.

3Step 3: Use Proposition 2.1.10 to Find the Limit

Proposition 2.1.10 (Monotone Convergence Theorem) states that if a sequence is monotone and bounded, then it converges. Since our sequence $ \frac{1}{\sqrt[3]{n}} $ is decreasing and bounded, it converges. As $ n \to \infty $, $ \sqrt[3]{n} \to \infty $. Thus, $ \frac{1}{\sqrt[3]{n}} \to 0 $. Therefore, the limit is 0.

Key Concepts

MonotonicityBounded SequencesLimitsMonotone Convergence Theorem

Monotonicity

Sequences can be monotonic in nature, meaning they are either entirely non-increasing or non-decreasing. Understanding monotonicity helps us predict the behavior of a sequence. In our example, the term given by the sequence is $ a_n = \frac{1}{\sqrt[3]{n}} $. We can explore whether $ a_n $ is decreasing by comparing it to its subsequent term, $ a_{n+1} $.
Doing the math shows:

For the sequence to be decreasing, we must have $ a_{n+1} \leq a_n $.
This translates to $ \frac{1}{\sqrt[3]{n+1}} \leq \frac{1}{\sqrt[3]{n}} $, which simplifies to $ n \leq n+1 $.
The inequality $ n \leq n+1 $ holds true for all natural numbers, so the sequence is indeed monotonically decreasing.

This monotonic decrease means each term of our sequence gets progressively smaller, a critical observation for further analysis.

Bounded Sequences

A sequence is bounded if it is confined within certain limits, meaning there exists both an upper and a lower bound. For the sequence $ a_n = \frac{1}{\sqrt[3]{n}} $, bounding it is straightforward once we know the nature of its elements.

Every sequential term is positive because for all $ n > 0 $, $ \sqrt[3]{n} > 0 $ and thus $ a_n > 0 $.
We notice $ a_1 = \frac{1}{\sqrt[3]{1}} = 1 $ which serves as a starting upper bound.
Since the sequence is monotonically decreasing, all subsequent terms are equally or even smaller than $ a_1 $, confirming $ a_n \leq 1 $.

With zero as an obvious lower boundary and one for an upper boundary, this sequence is neatly bounded. Being bounded is crucial because it leads to predictable outcomes, such as convergence.

Limits

The concept of a limit involves values a sequence approaches as terms progress to infinity. Calculating limits helps understand the ultimate outcome of sequences, like $ a_n = \frac{1}{\sqrt[3]{n}} $. First, notice how dramatic changes of $ n $ affect the sequence.

As $ n $ increases towards infinity, $ \sqrt[3]{n} $ similarly approaches infinity.
Therefore, the term $ \frac{1}{\sqrt[3]{n}} $ tends towards zero, shrinking closer and closer.

This behavior, derived through boundaries and monotonicity arguments, establishes the sequence's limit as 0. Limits are fundamental in analysis as they formalize how sequences behave at their extremities.

Monotone Convergence Theorem

The Monotone Convergence Theorem offers a powerful tool in real analysis. It asserts that a sequence that is both monotonic and bounded must converge. Applying this theorem to the sequence $ a_n = \frac{1}{\sqrt[3]{n}} $ clarifies its limit without complex calculations.

We've shown that our sequence is decreasing (monotonic) and bounded between 0 and 1.
According to the theorem, these two properties guarantee convergence.
The convergence leads us to a limit; as deduced earlier, $ a_n \to 0 $ as $ n \to \infty $.

The theorem is not only an assurance of convergence but also a guiding principle validating our deduction about the limit. It exemplifies why mastering foundational properties of sequences is key to simplifying complex analysis problems.

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