Problem 20 - Calculus: Early Transcendentals

Question

Calculate, to four decimal places, the first ten terms of the sequence and use them to plot the graph of the sequence by hand. Does the sequence appear to have a limit? If so, calculate it. If not, explain why. \( a_n = 2 + \frac {(-1)^n}{n} \)

Step-by-Step Solution

Answer

The sequence appears to have a limit, which is 2.

Step 1: Identify the Sequence Formula

The given sequence is defined by the formula \( a_n = 2 + \frac{(-1)^n}{n} \). Here, \( n \) is the term number in the sequence.

Step 2: Calculate the First Term

For \( n = 1 \), substitute into the formula: \[a_1 = 2 + \frac{(-1)^1}{1} = 2 - 1 = 1.\]So, the first term \( a_1 \) is 1.

Step 3: Calculate the Second Term

For \( n = 2 \), substitute into the formula: \[a_2 = 2 + \frac{(-1)^2}{2} = 2 + \frac{1}{2} = 2.5.\]So, the second term \( a_2 \) is 2.5.

Step 4: Calculate Additional Terms

Continue calculating terms using the sequence formula:- For \( n = 3 \): \[a_3 = 2 + \frac{(-1)^3}{3} = 2 - \frac{1}{3} = 1.6667.\]- For \( n = 4 \): \[a_4 = 2 + \frac{(-1)^4}{4} = 2 + \frac{1}{4} = 2.25.\]- For \( n = 5 \): \[a_5 = 2 + \frac{(-1)^5}{5} = 2 - \frac{1}{5} = 1.8.\]- For \( n = 6 \): \[a_6 = 2 + \frac{(-1)^6}{6} = 2 + \frac{1}{6} = 2.1667.\]- For \( n = 7 \): \[a_7 = 2 + \frac{(-1)^7}{7} = 2 - \frac{1}{7} = 1.8571.\]- For \( n = 8 \): \[a_8 = 2 + \frac{(-1)^8}{8} = 2 + \frac{1}{8} = 2.125.\]- For \( n = 9 \): \[a_9 = 2 + \frac{(-1)^9}{9} = 2 - \frac{1}{9} = 1.8889.\]- For \( n = 10 \): \[a_{10} = 2 + \frac{(-1)^{10}}{10} = 2 + \frac{1}{10} = 2.1.\]

Step 5: Plot the Sequence

Plot the ten calculated terms on a graph, using \( n \) as the horizontal axis and \( a_n \) as the vertical axis. You should see an oscillating pattern with the values alternating above and below 2.

Step 6: Analyze Limit Behavior

As \( n \) increases, the term \( \frac{(-1)^n}{n} \) approaches zero because the denominator \( n \) increases unboundedly. Hence, the sequence \( a_n = 2 + \frac{(-1)^n}{n} \) approaches 2. Thus, it appears the sequence has a limit, and the limit is 2. This is because the alternating term becomes negligible as \( n \) becomes large.

Key Concepts

Oscillating Sequences

An oscillating sequence is one that fluctuates between different values as it progresses. This gives it a characteristic pattern of alternating above and below a certain number. With the sequence formula given by \( a_n = 2 + \frac{(-1)^n}{n} \), it's important to note the presence of \((-1)^n\), which causes the terms to alternate between adding and subtracting the fraction \(\frac{1}{n}\).
This aspect of the sequence creates the oscillation effect, with consecutive terms moving towards and away from the number 2. As \( n \) increases, the size of the oscillation becomes smaller since the fraction \( \frac{1}{n} \) decreases. Thus, while the sequence continues to oscillate, the amplitude of this oscillation decreases, driving the terms closer to a specific value.
This concept is vital in understanding how certain sequences don't just increase or decrease, but instead keep shifting back and forth, usually converging over time.

Sequence Plotting

Plotting a sequence is a practical way to visualize its behavior. By using the sequence formula, in this case, \( a_n = 2 + \frac{(-1)^n}{n} \), we calculate the terms for different values of \( n \). We then plot these on a graph using \( n \) on the horizontal axis and \( a_n \) on the vertical axis.
For our sequence, we've calculated the first ten terms:

• \( a_1 = 1 \)
• \( a_2 = 2.5 \)
• \( a_3 = 1.6667 \)
• \( a_4 = 2.25 \)
• \( a_5 = 1.8 \)
• \( a_6 = 2.1667 \)
• \( a_7 = 1.8571 \)
• \( a_8 = 2.125 \)
• \( a_9 = 1.8889 \)
• \( a_{10} = 2.1 \)

When plotted, these points will reveal the oscillating nature of the sequence, with the pattern swinging above and below the number 2.
Visualizing the sequence helps students intuitively grasp how sequences behave and whether they lean towards a specific limit.

Convergence of Sequences

The concept of convergence in sequences describes how a sequence of numbers approaches a specific value. For our sequence, \( a_n = 2 + \frac{(-1)^n}{n} \), convergence means that as \( n \) becomes very large, \( a_n \) gets incredibly close to a certain number.
In this sequence, the term \( \frac{(-1)^n}{n} \) becomes negligible as \( n \) increases, since the denominator \( n \) grows infinitely, driving the value towards zero. Thus, \( a_n \) approaches the number 2, which is the limit of the sequence.
This happens because the oscillating addition (or subtraction) of \( \frac{1}{n} \) diminishes in impact as \( n \) gets larger, allowing \( a_n \) to settle more steadily around 2. Understanding convergence in this sequence assists students in predicting the eventual behavior of not just mathematical sequences, but related series and functions.