Problem 4 - Analysis 1: Ein Lehr- und Arbeitsbuch für Studienanfänger

Question

Prüfen Sie nach, ob die Folge \(\left(a_{n}\right), a_{n}=2^{n}\) aus Aufgabe 1 monoton wachsend oder monoton fallend ist.

Step-by-Step Solution

Answer

The sequence \( (a_n) = 2^n \) is monotonically increasing.

Step 1: Define Monotonically Increasing and Decreasing

A sequence \( (a_n) \) is monotonically increasing if for every \( n \), \( a_{n+1} \geq a_n \). It is monotonically decreasing if \( a_{n+1} \leq a_n \).

Step 2: Write the Sequence Formula

The sequence given is \( a_n = 2^n \), where each term \( a_n \) is a power of 2 based on \( n \).

Step 3: Compare Consecutive Terms

We need to compare \( a_{n+1} \) and \( a_n \). Calculate: \[ a_{n+1} = 2^{n+1} = 2 \cdot 2^n = 2a_n \] Clearly, \( a_{n+1} = 2a_n \).

Step 4: Determine Monotonic Behavior

Since \( a_{n+1} = 2a_n \), and because 2 is greater than 1, each term is double its preceding term. This shows that the sequence is monotonically increasing.

Key Concepts

Monotonically Increasing

A sequence is considered monotonically increasing if each term in the sequence is greater than or equal to the term before it. This means that as you move from one term to the next in the sequence, the value doesn’t decrease.

Let's focus on our sequence:

• If given a sequence \( a_n = 2^n \), we always check whether each subsequent term, \( a_{n+1} \), is greater than or the same as \( a_n \).
• For example, in our case, \( a_{n+1} = 2 \times a_n \), which simplifies to saying that the next term is twice the previous term.

Since multiplication by 2 scales up the value, each term is indeed larger than the previous one, establishing that the sequence is monotonically increasing, as it consistently increases with each new term.

Mathematical Proof

Mathematical proof is a logical argument that verifies the truth of a mathematical statement. For sequences, proving monotonicity involves showing a consistent relationship between consecutive terms.

In our sequence \( a_n = 2^n \):

• First, we establish the base case: \( a_1 = 2^1 = 2 \), which serves as our starting point and the smallest term with \( n = 1 \).
• Next, consider any \( n \): we are tasked to confirm if \( a_{n+1} \geq a_n \).
• Transforming \( a_{n+1} = 2^{n+1} = 2 \times 2^n = 2a_n \), clearly shows that \( a_{n+1} \) is twice \( a_n \).

Because multiplying by 2 (a positive constant greater than 1) undeniably increases the value, the proof confirms the sequence is monotonically increasing for all natural numbers \( n \).

Sequence Analysis

Sequence analysis involves examining the properties and behavior of sequences, which can include their growth patterns, limits, and whether they converge or diverge.

For a sequence like \( a_n = 2^n \), we are interested in:

• The growth pattern: Each term in the sequence is an exponential function of \( n \), which grows rapidly.
• The rate of increase: As confirmed, the sequence "doubles" each term, which means it exhibits exponential growth.
• Understanding potential limits: Since \( 2^n \) increases without bound as \( n \) approaches infinity, the sequence diverges, meaning it does not settle at any particular finite value.

Through analyzing the sequence \( a_n = 2^n \), we can fully recognize its characteristic of diverging to infinity, while maintaining a consistent and predictable growth pattern.