Problem 40 - Essential Calculus Early Transcendentals

Question

\(37-40=\) Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? $$a_{n}=n+\frac{1}{n}$$

Step-by-Step Solution

Answer

The sequence is increasing and bounded below by 2, but not above.

Step 1: Determine Monotonicity

To check if the sequence is increasing or decreasing, we will calculate the difference between consecutive terms, i.e. find \(a_{n+1} - a_{n}\). First, write the terms:\[a_n = n + \frac{1}{n}\] and \[a_{n+1} = (n+1) + \frac{1}{n+1}.\] The difference is: \[a_{n+1} - a_n = \left((n+1) + \frac{1}{n+1}\right) - \left(n + \frac{1}{n}\right) = 1 + \frac{1}{n+1} - \frac{1}{n}.\] Simplify the expression:\[1 + \frac{1}{n+1} - \frac{1}{n} = 1 + \left(\frac{1}{n+1} - \frac{1}{n}\right).\] The expression \(\frac{1}{n+1} - \frac{1}{n}\) is negative, hence the whole expression is greater than zero. Thus, the sequence is increasing.

Step 2: Check if the Sequence is Bounded

A sequence is bounded if there is a specific number that the sequence's terms cannot exceed for all \(n\). The sequence \(a_n = n + \frac{1}{n}\) has \(n\) as a term, which increases indefinitely, thus the sequence is not bounded above. To determine if it is bounded below, consider the smallest value it may take: when \(n=1\), \(a_1 = 1 + 1 = 2\). Hence, the sequence is bounded below by 2.

Key Concepts

sequence boundedness

In mathematics, a sequence is described as \'bounded\' if its terms do not surpass a certain fixed value. To be specific, a sequence is bounded from above if there exists a number, say \( M \), such that every term in the sequence is less than or equal to \( M \). Similarly, it is bounded from below if there exists a number, say \( m \), such that every term is greater than or equal to \( m \).
In the exercise, the sequence given is \( a_n = n + \frac{1}{n} \). Let’s analyze its boundedness:

• For upper boundedness, observe that as \( n \) tends to infinity, \( n + \frac{1}{n} \) increasingly grows without restrictions. This implies the sequence is not bounded from above.
• For lower boundedness, the smallest value of \( a_n \) is at \( n = 1 \), giving \( a_1 = 2 \). Thus, the sequence can be said to be bounded below by 2 as no term will ever go below this value.

So, the sequence \( a_n = n + \frac{1}{n} \) is not bounded above but is bounded below by 2.

sequence increasing

A sequence is classified as \'increasing\' if each term is greater than or equal to the previous one. Mathematically, this requires that \( a_{n+1} \geq a_n \) for every \( n \). To test this for the given sequence \( a_n = n + \frac{1}{n} \), we calculate the difference between consecutive terms:
We know:

• \( a_n = n + \frac{1}{n} \)
• \( a_{n+1} = (n+1) + \frac{1}{n+1} \)

The difference, \( a_{n+1} - a_n = 1 + \left(\frac{1}{n+1} - \frac{1}{n}\right) \), simplifies further to a positive value because \( \frac{1}{n+1} < \frac{1}{n} \). This indicates that each succeeding term is larger than the preceding one. Consequently, \( a_{n+1} - a_n > 0 \), traces that the sequence is indeed increasing. Thus, the sequence \( a_n = n + \frac{1}{n} \) is strictly increasing.

sequence analysis

Understanding sequences involves examining their behavior, which is the core of sequence analysis. We assess features like monotonicity, boundedness, convergence, or divergence.
In this problem, the sequence \( a_n = n + \frac{1}{n} \) displays certain characteristics:

• **Monotonicity**: The sequence is increasing if every term is greater than the previous term. This was demonstrated by assessing the difference between consecutive terms, showing that \( a_{n+1} > a_n \).
• **Boundedness**: Although we found that \( a_n \) is not bounded above because \( n \) grows indefinitely, it remains bounded below by a specific minimum value of 2.

Effective sequence analysis allows us to predict the long-term behavior of the sequence. Here, as \( n \) increases, \( a_n \) becomes infinitely large but never dips below 2. Such insights are crucial for deeper mathematical investigations.