Problem 5 - Calculus: Early Transcendentals

Question

List the first five terms of the sequence. \( a_n = \frac {(-1)^{n-1}}{5^n} \)

Step-by-Step Solution

Answer

The first five terms are \( \frac{1}{5} \), \( -\frac{1}{25} \), \( \frac{1}{125} \), \( -\frac{1}{625} \), \( \frac{1}{3125} \).

Step 1: Understand the Sequence Formula

The general term of the sequence is given by \( a_n = \frac {(-1)^{n-1}}{5^n} \). This formula involves two parts: a sign determined by \((-1)^{n-1}\) and a denominator of \(5^n\).

Step 2: Calculate the First Term

To find the first term \(a_1\), substitute \(n = 1\) into the formula: \[ a_1 = \frac{(-1)^{1-1}}{5^1} = \frac{(-1)^0}{5} = \frac{1}{5} \].

Step 3: Calculate the Second Term

To find the second term \(a_2\), substitute \(n = 2\) into the formula: \[ a_2 = \frac{(-1)^{2-1}}{5^2} = \frac{(-1)^1}{25} = -\frac{1}{25} \].

Step 4: Calculate the Third Term

To find the third term \(a_3\), substitute \(n = 3\) into the formula: \[ a_3 = \frac{(-1)^{3-1}}{5^3} = \frac{(-1)^2}{125} = \frac{1}{125} \].

Step 5: Calculate the Fourth Term

To find the fourth term \(a_4\), substitute \(n = 4\) into the formula: \[ a_4 = \frac{(-1)^{4-1}}{5^4} = \frac{(-1)^3}{625} = -\frac{1}{625} \].

Step 6: Calculate the Fifth Term

To find the fifth term \(a_5\), substitute \(n = 5\) into the formula: \[ a_5 = \frac{(-1)^{5-1}}{5^5} = \frac{(-1)^4}{3125} = \frac{1}{3125} \].

Key Concepts

Alternating Sequence

An alternating sequence is a type of mathematical sequence where the terms consistently switch between positive and negative signs.
This switching pattern often helps in assessing convergence in sequences and series. In our sequence formula, the expression \((-1)^{n-1}\) is responsible for the sign change in each term.

• When \( n \) is odd, \((-1)^{n-1}\) results in a positive value, specifically \(1\).
• When \( n \) is even, \((-1)^{n-1}\) results in a negative value, specifically \(-1\).

This alternating sign pattern can be visualized by examining the calculated terms:

• First term \( a_1 = \frac{1}{5} \)
• Second term \( a_2 = -\frac{1}{25} \)
• Third term \( a_3 = \frac{1}{125} \)
• Fourth term \( a_4 = -\frac{1}{625} \)
• Fifth term \( a_5 = \frac{1}{3125} \)

Each step, the power of \(5\) in the denominator increases, while the alternation of signs is controlled by \((-1)^{n-1}\). This creates a neat back-and-forth pattern, illustrating the core concept of an alternating sequence.

Exponential Sequence

Exponential sequences are characterized by a constant base raised to a variable exponent.
Here, the variable exponent is the term's position in the sequence, \( n \). This results in a consistent pattern of exponential growth or decay.

In this sequence, our constant base is \(5\), and the exponent is the term index \( n \). This is represented as \(5^n\), which serves as the denominator in our formula.

• When \( n = 1 \), the denominator becomes \( 5^1 = 5 \)
• When \( n = 2 \), the denominator becomes \( 5^2 = 25 \)
• When \( n = 3 \), the denominator becomes \( 5^3 = 125 \)
• Continuing this pattern results in rapid increase of denominators like \( 625 \) and \( 3125 \) for terms \( n = 4 \) and \( n = 5 \).

Exponential growth of the denominator causes the terms of the sequence to approach zero, illustrating how exponential sequences can contribute to sequences that converge towards a specific limit as \( n \) increases.

Sequence Terms Calculation

Calculating the terms of a sequence involves substituting specific values of \( n \) into the given formula.
Each result provides the exact numerical term of the sequence. The process requires a careful combination of factoring in the alternating sign caused by \((-1)^{n-1}\), and the growing denominator \(5^n\).

Let’s break it down step-by-step using the sequence formula \( a_n = \frac{(-1)^{n-1}}{5^n} \):

1. Identify whether the term is positive or negative by evaluating \((-1)^{n-1}\). For example:
   • For \( n=3 \), \((-1)^{3-1} = (-1)^2 = 1 \) (since 2 is even, the value is positive).
2. Calculate the denominator using \(5^n\). For example:
   • For \( n=3 \), calculate \( 5^3 = 125 \).
3. Combine these components to find the term \( a_3 = \frac{1}{125} \).

This systematic approach ensures that each term is computed accurately, reflecting both the alternating nature and the exponential decrease due to the larger denominator.