Question

For each of the following sequences, whose \(n\) th terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing. $$ n^{1 / n}, n \geq 3 $$

Short answer

The sequence is bounded below by 1 and eventually decreasing.

Step-by-step solution

Examine Boundedness

To check if the sequence given by the term \(a_n = n^{1/n}\) is bounded, examine the behavior of the exponent \(1/n\) as \(n\) increases. Notice that \(n^{1/n} = e^{\ln(n)/n}\). As \(n\) goes to infinity, \(\ln(n)/n\) tends to zero. Therefore, \(n^{1/n}\) approaches \(e^0 = 1\). Notice that for all \(n \geq 3\), \(n^{1/n}\) is greater than 1 and approaches 1, so it is bounded below by 1.

Check Monotonicity

To determine if the sequence \(n^{1/n}\) is eventually increasing or decreasing, analyze the difference \(a_{n+1} - a_n\). Sequence \(n^{1/n}\) is decreasing if \(\ln a_n > \ln a_{n+1}\). This simplifies to \(\frac{1}{n}\ln(n)<\frac{1}{n+1}\ln(n+1)\). For large \(n\), \(\ln(1+1/n) < \ln n - \ln(n+1)\) holds true. Therefore, for sufficiently large \(n\), the sequence is decreasing.

Numerical Approach for Small n

List a few terms of the sequence to observe the behavior or support theoretical findings. For example, compute \(3^{1/3} \approx 1.442\), \(4^{1/4} \approx 1.414\), etc. This confirms that as \(n\) increases, \(n^{1/n} \) indeed approaches but never falls below 1, and the terms decrease.

Key concepts

Bounded Sequence

In calculus, a sequence is said to be bounded if there is a real number that serves as an upper bound and another as a lower bound for all the terms in the sequence. For the sequence given by the term \( a_n = n^{1/n} \), we analyze its behavior as \( n \) increases, particularly focusing on larger values of \( n \). As \( n \) grows, the exponent \( 1/n \) decreases towards zero, which implies that the sequence approaches \( e^0 = 1 \). This reveals that the sequence cannot exceed 1 for large \( n \).
Not only does the sequence approach 1, but for all \( n \geq 3 \), \( n^{1/n} \) is actually greater than 1. This establishes 1 as a lower bound. Hence, we can conclude that the sequence is bounded below by 1.
Observing initial terms confirms that \( n^{1/n} \) begins at values greater than 1, exemplified by calculating \( 3^{1/3} \approx 1.442 \) and \( 4^{1/4} \approx 1.414 \). Therefore, the sequence is bounded between its smaller values and 1, confirming its bounded nature.

Monotonic Sequence

A sequence is monotonic if its terms consistently increase or decrease. To determine the behavior of our sequence \( a_n = n^{1/n} \), we must investigate whether it is consistently increasing or consistently decreasing.
By analyzing it more in-depth, we examine the difference between consecutive terms \( a_{n+1} \) and \( a_n \). The sequence is considered decreasing if \( a_{n+1} < a_n \). The logarithmic form of the terms facilitates this verification: \( \ln a_n = \frac{\ln n}{n} \) should be greater than \( \ln a_{n+1} = \frac{\ln(n+1)}{n+1} \) for the sequence to decrease.
Through simplification, this results in \( \ln(1 + 1/n) < \ln n - \ln(n+1) \) which indeed holds true for sufficiently large \( n \). Thus, the sequence is eventually decreasing as \( n \) increases beyond small initial values.
Therefore, while \( n^{1/n} \) is not decreasing for all small \( n \) (as earlier terms are greater than later ones), it becomes decreasing as \( n \) becomes rather large.

Convergence of Sequences

A sequence converges when it approaches a specific value as \( n \) tends to infinity. We say a sequence \( a_n \) converges to the limit \( L \) if, for every tiny positive number \( \epsilon \), there exists a natural number \( N \) such that for all \( n > N \), the absolute difference \( |a_n - L| < \epsilon \).
The sequence \( n^{1/n} \) vividly demonstrates convergence because as \( n \) increases, \( n^{1/n} \) approaches 1. This is mathematically evidenced as \( \ln(n)/n \) tends toward zero when \( n \) becomes very large, meaning \( n^{1/n} \) approaches \( e^0 = 1 \).
The convergence here is both empirical, through observing calculated sequence terms (like \( 3^{1/3} \approx 1.442 \), \( 4^{1/4} \approx 1.414 \)), and analytical, by proving mathematically that for any \( \epsilon > 0 \), \( n^{1/n} \) gets closer to 1 as \( n \) increases, fulfilling the fundamental definition of convergence.