Is the sequence defined by $a_n = (-1)^n$ convergent or divergent?
The sequence is defined by the formula \(a_n = (-1)^n\) for \(n = 1, 2, 3, \dots\). To analyze its behavior, we can list the first few terms:
\(a_1 = (-1)^1 = -1\)
\(a_2 = (-1)^2 = 1\)
\(a_3 = (-1)^3 = -1\)
\(a_4 = (-1)^4 = 1\)
The terms continue to oscillate indefinitely between \(-1\) and \(1\).
A sequence \((a_n)\) is said to be convergent if there exists a real number \(L\) such that for any positive number \(\epsilon > 0\), there exists a positive integer \(N\) such that for all \(n > N\), the condition \(|a_n - L| < \epsilon\) is satisfied. If no such limit \(L\) exists, the sequence is called divergent.
We examine whether the sequence \(a_n = (-1)^n\) approaches a single limit \(L\). As \(n\) grows larger, the sequence alternates between \(-1\) and \(1\). It does not stay arbitrarily close to any single number \(L\). If we choose a small \(\epsilon\) (e.g., \(\epsilon = 0.5\)), we cannot find a value \(N\) such that all subsequent terms are within \(\epsilon\) of a single number \(L\), because the terms always jump between \(-1\) and \(1\). Thus, the sequence does not converge to a single value.
Since the sequence \(a_n = (-1)^n\) does not approach any single real limit \(L\), it meets the definition of a divergent sequence. Therefore, the correct classification is that the sequence is divergent. This corresponds to option (b).
In real-world applications, oscillating sequences can represent physical phenomena such as alternating current (AC) or simple harmonic motion. Just as this sequence alternates between two values, an AC signal alternates its polarity periodically. Understanding whether a system settles into a steady state (converges) or continues to oscillate (diverges) is crucial for engineers analyzing the stability of electrical circuits or mechanical systems.