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Magazine Chapter 11: Problem 18
Determine the boundedness and monotonicity of the sequence with \(a_{n},\) as indicated. $$\frac{2^{n}}{4^{n}+1}$$.
Short Answer
Expert verified
The given sequence \(a_n = \frac{2^n}{4^n + 1}\) is bounded between 0 and 1 and is monotonically decreasing.
Step by step solution
01
Simplify the expression
First, let's simplify the expression. Notice that we can rewrite \(4^{n}\) as \({(2^{2})}^{n} = 2^{2n}\). Then the expression becomes
$$a_n = \frac{2^n}{2^{2n} + 1}$$
02
Determine the behavior of sequence as n increases
Now, let's analyze the behavior of the expression as n increases. To do this, let's find the limit of the expression as n goes to infinity:
$$\lim_{n \to \infty} \frac{2^n}{2^{2n} + 1}$$
We can simplify this by dividing both the numerator and denominator by \(2^n\):
$$\lim_{n \to \infty} \frac{1}{2^n + \frac{1}{2^n}}$$
As n goes to infinity, \(2^n\) grows infinitely large, resulting in the terms \(\frac{1}{2^n}\) approaching zero:
$$\lim_{n \to \infty} \frac{1}{2^n + \frac{1}{2^n}} = \frac{1}{\infty + 0} = 0$$
Since the limit of the expression approaches zero, we can conclude that the sequence decreases as n increases.
03
Determine boundedness
Now let's determine if the sequence is bounded. We already determined that the sequence decreases as n increases, meaning the terms get smaller with each increase in n. We also found that the limit of the expression is zero, meaning it approaches zero but never actually reaches it.
So for every term in the sequence, it must be greater than zero, i.e., \(a_n > 0\). Next, we need to find an upper bound. Since the numerator is always smaller than the denominator, the value of the expression will always be less than 1:
$$0 < a_n < 1$$
We've now shown that the sequence is always greater than 0 and less than 1, so it is bounded between 0 and 1.
04
Determine monotonicity
We found that the limit of the expression is 0 when n approaches infinity. This means that the terms in the sequence decrease as the index n increases. Therefore, we can conclude that the sequence is monotonically decreasing.
05
Conclusion
Based on our analysis, the given sequence
$$a_n = \frac{2^{n}}{4^{n} + 1}$$
is bounded between 0 and 1 and is monotonically decreasing.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of a Sequence
Understanding the limit of a sequence is crucial in analyzing its behavior as the sequence progresses towards infinity. The limit describes what the sequence's terms are approaching as the index, usually denoted by n, becomes very large.
For the sequence given by \(a_{n} = \frac{2^{n}}{4^{n} + 1}\), we seek to find \(\lim_{n \to \infty} a_{n}\). This exploration requires us to simplify our expression and find the behavior of the sequence as n increases. Through the simplification process, the original complex expression is reduced to a more manageable form by dividing both the numerator and the denominator by \(2^{n}\), leading us to \(\frac{1}{2^{n} + \frac{1}{2^{n}}}\).
As n approaches infinity, the term \(2^{n}\) becomes very large, making its reciprocal approach zero. The limit of the sequence is hence found to be zero. This notion of a limit is a foundational concept in calculus and helps us determine the long-term behavior of sequences and functions.
For the sequence given by \(a_{n} = \frac{2^{n}}{4^{n} + 1}\), we seek to find \(\lim_{n \to \infty} a_{n}\). This exploration requires us to simplify our expression and find the behavior of the sequence as n increases. Through the simplification process, the original complex expression is reduced to a more manageable form by dividing both the numerator and the denominator by \(2^{n}\), leading us to \(\frac{1}{2^{n} + \frac{1}{2^{n}}}\).
As n approaches infinity, the term \(2^{n}\) becomes very large, making its reciprocal approach zero. The limit of the sequence is hence found to be zero. This notion of a limit is a foundational concept in calculus and helps us determine the long-term behavior of sequences and functions.
Monotonically Decreasing Sequence
A sequence is considered to be monotonically decreasing if each term is less than or equal to the preceding term. In other words, as n increases, the value of each subsequent term in the sequence decreases or stays the same.
In our example \(a_{n} = \frac{2^{n}}{4^{n} + 1}\), we established that the limit as n approaches infinity is 0. This information alone is a strong indication that our sequence might be decreasing. To confirm this, we further observe that the denominator of our sequence's term \(4^{n} + 1\), which expands to \(2^{2n} + 1\), grows faster than the numerator \(2^{n}\). This result confirms that our sequence \(a_{n}\) is monotonically decreasing, as each increase in n results in a smaller fraction.
To make this concept more digestible, imagine a staircase where each step downwards represents an increase in n; the steps (terms of the sequence) get lower and lower, never rising back up.
In our example \(a_{n} = \frac{2^{n}}{4^{n} + 1}\), we established that the limit as n approaches infinity is 0. This information alone is a strong indication that our sequence might be decreasing. To confirm this, we further observe that the denominator of our sequence's term \(4^{n} + 1\), which expands to \(2^{2n} + 1\), grows faster than the numerator \(2^{n}\). This result confirms that our sequence \(a_{n}\) is monotonically decreasing, as each increase in n results in a smaller fraction.
To make this concept more digestible, imagine a staircase where each step downwards represents an increase in n; the steps (terms of the sequence) get lower and lower, never rising back up.
Bounded Sequence
A sequence is said to be bounded if there exists real numbers that serve as its upper and lower limits, within which all the terms of the sequence will fall. These bounds ensure that the sequence does not 'explode' to infinity or 'implode' to negative infinity.
In dealing with the sequence \(a_{n} = \frac{2^{n}}{4^{n} + 1}\), we have already concluded that it is monotonically decreasing and approaches a limit of 0. However, establishing that the sequence has a limit does not necessarily guarantee boundedness - we must show that there are explicit boundaries it cannot cross. The terms of this particular sequence are always positive, providing us with a natural lower bound of 0. Furthermore, through algebraic observation, we can see that the sequence's terms are always less than 1, because the numerator is smaller than the denominator. Therefore, we have shown that \(0 < a_{n} < 1\) for all terms in the sequence, proving that it is indeed a bounded sequence. This concept is paramount in understanding constraints within sequences and is a fundamental part of sequence analysis.
In dealing with the sequence \(a_{n} = \frac{2^{n}}{4^{n} + 1}\), we have already concluded that it is monotonically decreasing and approaches a limit of 0. However, establishing that the sequence has a limit does not necessarily guarantee boundedness - we must show that there are explicit boundaries it cannot cross. The terms of this particular sequence are always positive, providing us with a natural lower bound of 0. Furthermore, through algebraic observation, we can see that the sequence's terms are always less than 1, because the numerator is smaller than the denominator. Therefore, we have shown that \(0 < a_{n} < 1\) for all terms in the sequence, proving that it is indeed a bounded sequence. This concept is paramount in understanding constraints within sequences and is a fundamental part of sequence analysis.
