Problem 48
Question
In Exercises 39-48, write the first five terms of the sequence and find the limit of the sequence (if it exists). If the limit does not exist, explain why. Assume \(n\) begins with 1. $$ a_n = \dfrac{(-1)^{n+1}}{n^2} $$
Step-by-Step Solution
Verified Answer
The first five terms of the sequence are 1, -1/4, 1/9, -1/16, and 1/25. The limit of the sequence as \(n\) approaches infinity is 0.
1Step 1: Calculate the first five terms
The first step is to substitute the values \(n = 1, 2, 3, 4, 5\) into the formula for \(a_n\) to find the first five terms. For \(n = 1\), we get \(a_1 = \dfrac{(-1)^2}{1^2} = 1\). Similarly, for \(n = 2\), we find \(a_2 = \dfrac{(-1)^3}{2^2} = -\dfrac{1}{4}\). Repeat these steps for \(n = 3, 4, 5\).
2Step 2: Determine the limit of the sequence
To find the limit of the sequence as \(n\) approaches infinity, we can observe the behavior of the terms. As \(n\) becomes very large, the term \(-1^{n+1}\) continues to alternate between -1 and 1, but the denominator \(n^2\) becomes extremely large. As a result, the whole fraction approaches 0.
3Step 3: Conclusion
So, the first five terms of the sequence are 1, -1/4, 1/9, -1/16, and 1/25. The limit of the sequence as \(n\) approaches infinity is 0, because the denominator grows much faster than the numerator.
Key Concepts
Convergence of SequencesAlternating SeriesSeries Terms Calculation
Convergence of Sequences
When we talk about convergence of sequences, we are looking to answer one fundamental question: as the terms in the sequence progress (usually towards infinity), do they settle towards a specific value? This specific value, if it exists, is known as the limit of the sequence. The given exercise \( a_n = \frac{(-1)^{n+1}}{n^2} \) is a fantastic example to explore this concept.
For a sequence to converge, we must see that, as we pick larger and larger values of \(n\), the terms \(a_n\) get arbitrarily close to a particular number. If the terms do not approach any value and continue to be erratic, the sequence is said to diverge. It's essential to note that all convergent sequences are bounded, but not all bounded sequences converge. The provided step-by-step solution involves observing that as \(n\) increases indefinitely, the value of \(a_n\) approaches 0. This conclusion is pivotal since it indicates the presence of a limit and hence the convergence of the provided sequence.
For a sequence to converge, we must see that, as we pick larger and larger values of \(n\), the terms \(a_n\) get arbitrarily close to a particular number. If the terms do not approach any value and continue to be erratic, the sequence is said to diverge. It's essential to note that all convergent sequences are bounded, but not all bounded sequences converge. The provided step-by-step solution involves observing that as \(n\) increases indefinitely, the value of \(a_n\) approaches 0. This conclusion is pivotal since it indicates the presence of a limit and hence the convergence of the provided sequence.
Alternating Series
An alternating series is one in which the signs of the terms switch back and forth between positive and negative. This characteristic is typically represented mathematically by a factor of \( (-1)^n \) or \( (-1)^{n+1} \) in the series formula. In our exercise, \( a_n \) is an example of an alternating series because the \( (-1)^{n+1} \) term causes the sequence to alternate as \(n\) changes.
It's crucial to observe the alternating pattern because it affects the way we examine the convergence of a sequence. For an alternating sequence, if the magnitude of the terms (ignoring the sign) decreases toward zero, the series can be said to be convergent. This stems from an important result in mathematics called the Alternating Series Test. Our sequence satisfies this condition as the numerator alternates signs while the denominator \( n^2 \) grows without bound, ensuring the terms get closer to 0 as \(n\) increases.
It's crucial to observe the alternating pattern because it affects the way we examine the convergence of a sequence. For an alternating sequence, if the magnitude of the terms (ignoring the sign) decreases toward zero, the series can be said to be convergent. This stems from an important result in mathematics called the Alternating Series Test. Our sequence satisfies this condition as the numerator alternates signs while the denominator \( n^2 \) grows without bound, ensuring the terms get closer to 0 as \(n\) increases.
Series Terms Calculation
The process of series terms calculation is central to understanding how sequences behave. It requires substituting different values of \(n\) into the sequence formula to obtain the first few terms, which can expose patterns and insights about the sequence's nature. The exercise provided starts by calculating the first five terms, which is often a good practice for visualizing the sequence's progression.
When calculating terms of an alternating sequence especially, it is critical to pay attention not just to the magnitude of each term but also its sign. This determines the sequence’s behavior and helps to answer whether it converges or diverges. In our case, this careful calculation brings to light that the series alternates between positive and negative values, and that the magnitude decreases in a manner that suggests convergence. The simplicity of the formula for \(a_n\) allows us to see this pattern clearly, leading us to the conclusion that the limit is 0.
When calculating terms of an alternating sequence especially, it is critical to pay attention not just to the magnitude of each term but also its sign. This determines the sequence’s behavior and helps to answer whether it converges or diverges. In our case, this careful calculation brings to light that the series alternates between positive and negative values, and that the magnitude decreases in a manner that suggests convergence. The simplicity of the formula for \(a_n\) allows us to see this pattern clearly, leading us to the conclusion that the limit is 0.
Other exercises in this chapter
Problem 47
In Exercises 47 and 48, find $$\textrm{(a)}\ \lim_{x \to 2}\ f(x), \quad \textrm{(b)} \lim_{x \to 2}\ g(x), \quad \textrm{(c)} \lim_{x \to 2}\ [f(x)g(x)], \quad
View solution Problem 48
In Exercises 37-48, use the limit process to find the area of the region between the graph of the function and the x-axis over the specified interval. $$ f(x) =
View solution Problem 48
In Exercises 43-50, (a) find the slope of the graph of \(f\) at the given point, (b) use the result of part (a) to find an equation of the tangent line to the g
View solution Problem 48
In Exercises 37-48, use a graphing utility to graph the function and approximate the limit accurate to three decimal places. $$\lim_{x \to 0} (1+2x)^{1/x}$$
View solution