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Determine whether the sequence converges or diverges. $$a_{n}=\cos \pi n$$

Short Answer

Expert verified
The sequence diverges.

Step by step solution

01

Understand the Sequence

First, we need to acknowledge what the sequence looks like. We can take a few terms of the sequence: when \(n = 1, a_{1} = \cos(\pi) = -1\), when \(n = 2, a_{2} = \cos(2\pi) = 1\), when \(n = 3, a_{3} = \cos(3\pi) = -1\) and so on. Thus we see that for odd \(n\), \(a_{n} = -1\) and for even \(n\), \(a_{n} = 1\). So the sequence looks like this: -1, 1, -1, 1, -1, 1, ...
02

Apply Definition of Convergence

A sequence is said to converge if its terms get closer and closer to a particular value as one goes deeper into the sequence. However, if we inspect our sequence \(a_{n} = \cos \pi n\), we can see that its terms alternate between -1 and 1 without getting closer to any specific value. Therefore, this sequence does not converge to a particular number.
03

Final Conclusion

Since the sequence does not converge to any particular number, according to the definition, we can conclude that the sequence \(a_{n}= \cos \pi n\) diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sequence Divergence
When we talk about sequence divergence, we refer to the behavior of a sequence whose terms do not approach a single, specific value as the sequence progresses. Instead, the terms may either keep increasing or decreasing without bound, or they may oscillate without settling down to a limit.

In our exercise involving the sequence \(a_{n} = \cos(\pi n)\), the sequence's terms alternate between -1 and 1 for each subsequent value of \(n\) without approaching any particular number. This lack of approach to a specific value means that the sequence does not have a limit, and it exhibits a type of divergence.

To reinforce the concept, imagine trying to pinpoint a target on a dartboard: if your throws keep hitting wildly different points without ever getting consistently closer to the bullseye, you're essentially experiencing divergence in your aim.
Alternating Sequence
An alternating sequence is a special type of numeric sequence where the sign of the terms alternates between positive and negative. This behavior can be seen in sequences that involve a term like \( (-1)^{n} \) or a trigonometric function that induces alternation, like in our exercise with \( \cos(\pi n) \).

An introductory glance at an alternating sequence would show a pattern similar to -1, 1, -1, 1, ..., which precisely describes our sequence \( a_{n} \). Despite being a simple pattern, such a sequence challenges our intuitive understanding, as it doesn't converge in a traditional sense. This alternation is a clear signal for further inspection to determine if a sequence may have other forms of convergence or if it diverges, as in the given exercise.
Limit of a Sequence
The limit of a sequence is a foundational concept in calculus and analysis that captures the idea of where a sequence is headed as its terms progress further and further along. Mathematically, we say that a sequence \( \{a_{n}\} \) converges to a limit \( L \) if, for any chosen proximity around \( L \) (no matter how small), there exists a point in the sequence after which all subsequent terms fall within that proximity.

For a sequence to have a limit, the terms must get arbitrarily close to \( L \) as \( n \) gets large. If this condition is not met, as with our alternating sequence \( a_{n} = \cos(\pi n) \), then the sequence lacks a limit, making it divergent.

Visualizing Limits

If one were to graph the terms of a convergent sequence, they would see the points bunching up closer and closer to a single value on the number line as \( n \) increases. For a divergent sequence, the points do not cluster around any single value, indicating the absence of a limit.

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