Question

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit. $$ a_{n}=\ln \left(\frac{1}{n}\right) $$

Short answer

The sequence is unbounded, monotonic decreasing, and divergent to \(-\infty\).

Step-by-step solution

Analyze the Boundedness

A sequence is bounded if there exist real numbers \( M \) and \( m \) such that \( m \leq a_n \leq M \) for all \( n \). Since \( a_n = \ln\left(\frac{1}{n}\right) \) and \( n \geq 1 \), \( \frac{1}{n} \leq 1 \). Therefore, \( \ln\left(\frac{1}{n}\right) \leq 0 \). As \( n \to \infty \), \( \frac{1}{n} \to 0^+ \), hence \( \ln\left(\frac{1}{n}\right) \to -\infty \). Thus, the sequence is not bounded below as it can become arbitrarily large negative, i.e., \(-\infty < a_n \leq 0\).

Determine Monotonicity

A sequence is monotonic if it is entirely non-increasing or non-decreasing. Consider the function \( f(x) = \ln\left(\frac{1}{x}\right) \). Examine its derivative \( f'(x) = -\frac{1}{x} \), which is negative for all \( x > 0 \). This implies \( a_n \) is a decreasing function. Therefore, the sequence is monotonic decreasing.

Analyze Convergence or Divergence

A sequence converges if it approaches a finite limit as \( n \to \infty \). Since we have \( a_n = \ln\left(\frac{1}{n}\right) = -\ln(n) \), as \( n \to \infty \), \( \ln(n) \to \infty \) which means \( a_n \to -\infty \). Consequently, the sequence diverges.

Key concepts

Boundedness of Sequences

Sequences are mathematical lists of numbers that often follow a specific rule or pattern. For a sequence to be bounded, it means there is a real number limit, both above and below, that the sequence cannot surpass. In other words, a bounded sequence doesn't go beyond these set limits, whether positive or negative.
For the sequence given by \( a_n = \ln\left(\frac{1}{n}\right) \), assessing boundedness involves examining the behavior of the sequence as \( n \) changes. Since \( \frac{1}{n} \leq 1 \) for \( n \geq 1 \), we determine that \( \ln\left(\frac{1}{n}\right) \leq 0 \). Therefore, all terms are less than or equal to zero. However, observe as \( n \) becomes larger, \( \frac{1}{n} \) approaches zero. This shows \( \ln\left(\frac{1}{n}\right) \) decreases toward \(-\infty\), making the sequence unbounded below.
In simpler terms, while the sequence stays below zero, it has no lower limit due to its ability to decrease indefinitely.

Monotone Sequence

A monotone sequence is one that always moves in the same direction: it's either always increasing or always decreasing. Understanding this characteristic helps predict where the sequence is headed.
Our sequence \( a_n = \ln\left(\frac{1}{n}\right) \) exemplifies a decreasing sequence. To identify this, we can take a closer look through calculus by evaluating the derivative of the function \( f(x) = \ln\left(\frac{1}{x}\right) \). The derivative, \( f'(x) = -\frac{1}{x} \), is negative for all \( x > 0 \). A negative derivative indicates the sequence decreases as \( x \) (or \( n \)) increases.
Hence, each subsequent term in \( a_n \) is less than the preceding one. This clarifies that the sequence is monotonic and decreasing in nature.

Convergence and Divergence of Sequences

Convergence of a sequence refers to the sequence approaching a specific value or limit as the number of terms becomes infinitely large. If such a limit exists, the sequence is said to converge. Conversely, if no such finite limit is achieved, we state that the sequence diverges.
In the provided sequence \( a_n = \ln\left(\frac{1}{n}\right) \), as \( n \to \infty \), \( \frac{1}{n} \) tends towards zero. Therefore, \( \ln\left(\frac{1}{n}\right) = -\ln(n) \) transforms such that \( \ln(n) \to \infty \), leading \( a_n \to -\infty \).
Since the terms do not approach a finite value, the sequence diverges. Simply put, the sequence goes to negative infinity and doesn't settle around any certain number, confirming its divergent behavior.