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A factorial sequence is one where each term is defined as the factorial of a natural number. The factorial of a number, denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). In terms of sequences, when \( a_n = n! \), we can observe this sequence by calculating its initial elements:

• \( a_1 = 1! = 1 \)
• \( a_2 = 2! = 2 \)
• \( a_3 = 3! = 6 \)
• \( a_4 = 4! = 24 \)

Notice how each subsequent term is greater than the previous one. This is because multiplying a number by an integer greater than 1 guarantees an increase, which characterizes all terms in the factorial sequence as "strictly increasing." This property is crucial in monotonic sequences, wherein the sequence either consistently increases or decreases.

Exponential decay describes a process where quantities diminish at rates proportional to their current value. In a sequence like \( a_n = 2^{-n} \), the decay factor is the number 2 raised to the power of \(-n\). This particular sequence decreases because each subsequent term is half the previous term:

• \( a_1 = 2^{-1} = \frac{1}{2} \)
• \( a_2 = 2^{-2} = \frac{1}{4} \)
• \( a_3 = 2^{-3} = \frac{1}{8} \)
• \( a_4 = 2^{-4} = \frac{1}{16} \)

Here, we see that \( a_{n+1} \) is always less than \( a_n \), which indicates a "strictly decreasing" sequence. This type of behavior is the core attribute of sequences that undergo exponential decay, as each term is a smaller fraction of the preceding one.

Sequence comparison involves evaluating the relationship between terms in a sequence. For a sequence to be classified under a specific monotonic behavior, you must determine whether the sequence values consistently increase, decrease, or remain unchanged.
For instance, in comparing sequences where \( a_n = n! \) and \( a_n = n + \frac{1}{n} \), we recognize both are "strictly increasing." This is done by comparing \( a_{n+1} \) and \( a_n \) directly.
For \( a_n = n! \), knowing \( a_{n+1} \) equals \((n+1)! = (n+1) \cdot n!\), we see it is greater than \( n! \).
Meanwhile, in \( a_n = 2^{-n} \), calculating \( a_{n+1} \) and finding it equals \( \frac{1}{2} \ a_n \), shows it is less than \( a_n \). Sequence comparison is often the initial step to determine the overall pattern or trend within a sequence.

A sequence is strictly increasing if each term is larger than the one before it. This means that for any natural number \( n \), \( a_{n+1} > a_n \). This increase is "strict" because it holds without exception throughout.
Consider the sequence \( a_n = n + \frac{1}{n} \):

• Here, adding \( \frac{1}{n} \) ensures the terms don't simplify into just \( n \), yet they still grow with increasing values of \( n \).
• So, with \( a_1 = 1 + 1 = 2 \) and progressing, \( a_2 = 2 + \frac{1}{2} = 2.5 \) becomes greater than \( a_1 \).

This property of a strictly increasing sequence helps in making predictions and establishing bounds, as you know subsequent terms will exceed previous ones by at least some positive margin.
Strictness ensures that there are no plateaus or static terms.

A sequence is strictly decreasing when each term is lesser than the preceding one. For such sequences, \( a_{n+1} < a_n \) holds true for all terms, making the decrease consistent. An example is \( a_n = 2^{-n} \):

• Here, the values diminish by taking each prior term and multiplying it by \( \frac{1}{2} \), which is less than 1.
• Starting with \( a_1 = \frac{1}{2} \), this leads naturally to \( a_2 = \frac{1}{4} \), then \( a_3 = \frac{1}{8} \).

The sequence continues on a downward trajectory, visibly signifying reduction.
The strictness implies no two terms are equal, preserving a continuous downgrade in value. Such a categorization is helpful in identifying limits or concluding asymptotic behavior as it narrows towards zero or another lower boundary over time.