Question

Find \(\sup x_{n}, \inf x_{n}, \max x_{n},\) and \(\min x_{n}\) (if any) for sequences with general term (a) \(n ;\) (b) \((-1)^{n}\left(2-2^{2-n}\right)\); (c) \(1-\frac{2}{n} ;\) (d) \(\frac{n(n-1)}{(n+2)^{2}}\). Which are bounded in \(E^{1} ?\)

Short answer

Sequences (a) and (d) are unbounded; (b) and (c) are bounded in \(E^1\).

Step-by-step solution

Determine Supremum and Infimum for Sequence (a)

Sequence (a) has the general term given by the sequence \(x_n = n\). Since \(n\) increases without bound as \(n\) increases, \(\sup x_n = \infty\) and \(\inf x_n = 1\) if we start from \(n=1\). This sequence does not have a maximum or minimum as it is unbounded and always increasing.

Analyze Supremum and Infimum for Sequence (b)

For sequence (b), the general term is \((-1)^{n}(2-2^{2-n})\). Calculating a few terms: when \(n=1\), it is \(-1\cdot 0.5 = -0.5\); when \(n=2\), it is \(1\cdot 1 = 1\), when \(n=3\), it is \(-1\cdot 1.5 = -1.5\), and so forth. Observing the alternating nature, the maximum is \(1\) and minimum is \(-1.5\), hence \(\sup x_n = 1\) and \(\inf x_n = -1.5\). Max and min values are both realized within the sequence.

Derive Maximum and Minimum for Sequence (c)

Sequence (c) is given by \(x_n = 1 - \frac{2}{n}\). As \(n\) approaches infinity, \(\frac{2}{n}\) approaches zero, making \(x_n\) approach \(1\). Thus, \(\sup x_n = 1\) and the infimum is \(0\) for \(n=2\). The minimum is \(-1\) as \(\frac{2}{n}\) cannot equal \(2\), so no maximum or minimum values exist within the sequence itself.

Compute Bounds for Sequence (d)

Sequence (d) has the formula \(x_n = \frac{n(n-1)}{(n+2)^{2}}\). Using L'Hôpital's rule or observation, as \(n\) goes to infinity, \(x_n\) approaches a horizontal \(0\), indicating that \(\sup x_n\approx \frac{1}{4}\), calculated by plugging low \(n\) values, and \(\inf x_n=0\). There's no maximum or minimum within the sequence.

Check Boundedness in E^1

Sequences (a) and (d) are unbounded in \(E^1\) as (a) tends to infinity and (d) approaches \( \frac{1}{4} \) with no fixed bounds. Sequence (b) is bounded between \(-1.5\) and \(1\), while sequence (c) is bounded between \(0\) and \(1\). Hence, only sequences (b) and (c) are bounded.

Key concepts

Bounded Sequences

Understanding bounded sequences is crucial in sequence analysis. Simply put, a sequence is called bounded if there is a real number that acts as an upper bound and another as a lower bound for all terms in the sequence. This means all terms of the sequence stay within these boundaries.

If a sequence has an upper bound and a lower bound, it does not reach infinity in any direction. For example, observe Sequence (b) \((-1)^n (2 - 2^{2-n})\):

• This sequence is bounded because it oscillates between a maximum of 1 and a minimum of -1.5, thus constrained within finite limits.
• Similarly, Sequence (c) \(1 - \frac{2}{n}\) has boundaries, as its terms stay within 0 and 1.

However, sequences like (a) \(n\) and (d) \(\frac{n(n-1)}{(n+2)^2}\) are not bounded, as one does not have an upper limit and the other tends to infinity or a value without a definitive upper bound.

Understanding which sequences are bounded helps in evaluating their behavior over time and estimating their extremities like maximum and minimum values.

Maximum and Minimum

When studying sequences, identifying the maximum and minimum values often helps understand the extent or the range of the sequence values. These maxima or minima are individual sequence values that represent the strongest upper and lower limits within the sequence:

- **Maximum** is the largest value a sequence term can reach. - **Minimum** is the smallest value a sequence term can take.

When talking about sequence analysis, like in our example:

• Sequence (b) shows a distinct maximum at 1 and minimum at -1.5. These represent the tallest and the lowest points reached by the sequence terms.
• In contrast, sequence (c) does not have a distinct minimum or maximum within its terms, even though it stays bounded between 0 and 1.

Spotting maximums and minimums helps in quickly understanding sequences' limits and can guide predictions about their behavior through graphical or numeric methods.

Sequence Analysis

Sequence analysis is an integral part of understanding the characteristics and behavior of sequences over time. It involves examining the progression of sequence terms and identifying any particular features, such as trends or oscillations.

Through sequence analysis, students can deduce whether a sequence is convergent or divergent. Here's a closer look:

• Convergent sequences, like (c) \(1 - \frac{2}{n}\), approach a distinct limit or boundary as the number of terms increases.
• Divergent sequences, such as (a) \(n\), continuously increase without settling into a specific limit.

Sequence analysis can also reveal if sequences are periodic or have a repeating cycle, evidenced by the alternating pattern in sequence (b).

By analyzing sequences, students can uncover significant properties like - The rate at which terms increase or decrease,- Potential for periodicity,- Extent of oscillation in the sequence.These insights aid in other mathematical applications, including calculus and statistical analyses, providing a foundation for understanding functions and their behaviors as they translate into real-world applications.