Problem 25
Question
In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). \(1, \frac{2}{2^{2}-1^{2}}, \frac{3}{3^{2}-2^{2}}, \frac{4}{4^{2}-3^{2}}, \ldots\)
Step-by-Step Solution
Verified Answer
The explicit formula is \( a_n = \frac{n}{2n-1} \). The sequence converges with \( \lim_{n \to \infty} a_n = \frac{1}{2} \).
1Step 1: Identify the Pattern in the Sequence
The sequence provided is structured as \( 1, \frac{2}{2^2-1^2}, \frac{3}{3^2-2^2}, \frac{4}{4^2-3^2}, \ldots \). Let's try to find a common pattern in the terms \( a_n = \frac{n}{n^2-(n-1)^2} \). Each term numerator increases by 1 each time, starting at 1, and the denominator is the difference of squares: \( n^2-(n-1)^2 \).
2Step 2: Simplify the Denominator
The denominator \( n^2 - (n-1)^2 \) simplifies. Expanding \((n-1)^2\), we get \( n^2 - (n^2 - 2n + 1) = n^2 - n^2 + 2n - 1 = 2n - 1\). This simplifies the general term to \( a_n = \frac{n}{2n-1} \).
3Step 3: Determine Convergence or Divergence
Now, determine if the sequence converges or diverges. To check convergence, evaluate \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n}{2n-1} = \lim_{n \to \infty} \frac{1}{2 - \frac{1}{n}}\). As \( n \to \infty \), the term \( \frac{1}{n} \to 0 \), so \( \lim_{n \to \infty} \frac{1}{2 - \frac{1}{n}} = \frac{1}{2} \). Thus, the sequence converges.
Key Concepts
Explicit FormulaLimit of a SequenceDifference of Squares
Explicit Formula
An explicit formula is a way to express the terms of a sequence as a function of their position, denoted by the index variable, usually represented as \( n \). This allows us to find any term in the sequence without needing to know the previous terms. In mathematics, finding an explicit formula can make it significantly easier to assess the behavior of the sequence, especially as the terms grow large.
In the original exercise, we have a sequence: \( 1, \frac{2}{2^2-1^2}, \frac{3}{3^2-2^2}, \frac{4}{4^2-3^2}, \ldots \). By identifying the pattern through simplification, we determined an explicit formula for the \( n \)-th term, \( a_n = \frac{n}{n^2-(n-1)^2} \). After simplifying the denominator using the difference of squares, we get \( a_n = \frac{n}{2n - 1} \).
Using this formula, we can calculate any term without explicitly listing each one, which aids in understanding both the sequence and its behavior in the long run.
In the original exercise, we have a sequence: \( 1, \frac{2}{2^2-1^2}, \frac{3}{3^2-2^2}, \frac{4}{4^2-3^2}, \ldots \). By identifying the pattern through simplification, we determined an explicit formula for the \( n \)-th term, \( a_n = \frac{n}{n^2-(n-1)^2} \). After simplifying the denominator using the difference of squares, we get \( a_n = \frac{n}{2n - 1} \).
Using this formula, we can calculate any term without explicitly listing each one, which aids in understanding both the sequence and its behavior in the long run.
Limit of a Sequence
When working with sequences, the limit describes the behavior of the sequence as the index \( n \) approaches infinity. In other words, it helps us to understand what value, if any, the sequence gets closer to as \( n \) becomes very large.
To find the limit of our sequence, expressed by the formula \( a_n = \frac{n}{2n - 1} \), we examine the behavior of the sequence by evaluating \( \lim_{n \to \infty} a_n \). The steps to evaluate this limit include dividing both the numerator and the denominator by \( n \), resulting in \( \lim_{n \to \infty} \frac{1}{2 - \frac{1}{n}} \). As \( n \) grows larger, \( \frac{1}{n} \rightarrow 0 \).
This simplifies the limit to \( \frac{1}{2} \), indicating that the sequence converges to \( \frac{1}{2} \) as \( n \) approaches infinity.
To find the limit of our sequence, expressed by the formula \( a_n = \frac{n}{2n - 1} \), we examine the behavior of the sequence by evaluating \( \lim_{n \to \infty} a_n \). The steps to evaluate this limit include dividing both the numerator and the denominator by \( n \), resulting in \( \lim_{n \to \infty} \frac{1}{2 - \frac{1}{n}} \). As \( n \) grows larger, \( \frac{1}{n} \rightarrow 0 \).
This simplifies the limit to \( \frac{1}{2} \), indicating that the sequence converges to \( \frac{1}{2} \) as \( n \) approaches infinity.
- Useful in predicting long-term behavior.
- Convergence means the sequence approaches a specific value.
Difference of Squares
The difference of squares is a mathematical expression used in algebra to describe the situation where two squared terms are subtracted: \( a^2 - b^2 \). This can be factorized into \( (a-b)(a+b) \). By recognizing this pattern, many expressions can be simplified, making them easier to work with.
In the sequence provided in the problem, the denominator involves the difference of squares: \( n^2 - (n-1)^2 \). Upon simplifying, this becomes \( n^2 - (n^2 - 2n + 1) = 2n - 1 \), which reveals a much simpler form to work with.
Understanding and identifying the difference of squares:
In the sequence provided in the problem, the denominator involves the difference of squares: \( n^2 - (n-1)^2 \). Upon simplifying, this becomes \( n^2 - (n^2 - 2n + 1) = 2n - 1 \), which reveals a much simpler form to work with.
Understanding and identifying the difference of squares:
- Enables simplification of complex algebraic expressions.
- Is an essential skill for simplifying sequences or expressions.
- Facilitates the discovery of explicit formulas and limits.
Other exercises in this chapter
Problem 24
Show that \(\sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right)=-\ln 2\).
View solution Problem 25
In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ 1+\frac
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In Problems 13–30, classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{\sin n}{n \sqrt{n
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Find the sum of each of the following series by recognizing how it is related to something familiar. (a) \(x-x^{2}+x^{3}-x^{4}+x^{5}-\cdots\) (b) \(\frac{1}{2 !
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