Problem 38
Question
In Exercises \(35-40,\) find a formula for the \(n\) th partial sum of the series and use it to determine if the series converges or diverges. If a series converges, find its sum. $$ \sum_{n=1}^{\infty}(\tan (n)-\tan (n-1)) $$
Step-by-Step Solution
Verified Answer
The series diverges as the tangent function does not converge as \( n \to \infty \).
1Step 1: Understand the structure of the series
The given series is \( \sum_{n=1}^{\infty}(\tan(n) - \tan(n-1)) \). This is a telescoping series, where terms typically cancel each other out. Our task is to find the formula for the \( n \)-th partial sum and analyze the remaining terms.
2Step 2: Write the terms of the series
To understand how this series behaves, write out the first few terms: \((\tan(1) - \tan(0)) + (\tan(2) - \tan(1)) + (\tan(3) - \tan(2)) + \ldots\). Observe that most terms are canceling each other, leaving only \(\tan(n) - \tan(0)\) after the cancellation process.
3Step 3: Find the n-th partial sum
In a telescoping series, the nth partial sum \(S_n\) is given by the initial terms that do not get canceled. Thus, for this series, the \( n \)-th partial sum is \( S_n = \tan(n) - \tan(0) \). Since \( \tan(0) = 0 \), it simplifies to \( S_n = \tan(n) \).
4Step 4: Determine convergence or divergence
The series converges if the limit of the partial sums \( S_n \) as \( n \to \infty \) exists. Since \( S_n = \tan(n) \) and \( \tan(n) \) does not converge to a finite limit as \( n \to \infty \) (because the tangent function oscillates between \( -\infty \) and \( \infty \)), the series diverges.
Key Concepts
Partial SumsConvergence and Divergence of SeriesInfinite Series Analysis
Partial Sums
When dealing with series, partial sums are a crucial component. They represent the sum of the first several terms, up to the nth term, of a series. Considering our specific example of a telescoping series, each partial sum builds upon the cancellations of consecutive terms.
In the series \( \sum_{n=1}^{\infty}(\tan(n) - \tan(n-1)) \), the nth partial sum, denoted as \( S_n \), can be expressed explicitly as \( S_n = \tan(n) - \tan(0) \). Since \( \tan(0) = 0 \), this simplifies to \( S_n = \tan(n) \). The recent term essentially gives the main value for each partial sum in this telescoping arrangement.
In the series \( \sum_{n=1}^{\infty}(\tan(n) - \tan(n-1)) \), the nth partial sum, denoted as \( S_n \), can be expressed explicitly as \( S_n = \tan(n) - \tan(0) \). Since \( \tan(0) = 0 \), this simplifies to \( S_n = \tan(n) \). The recent term essentially gives the main value for each partial sum in this telescoping arrangement.
- Partial sums often help in observing the behavior of a series across its progression.
- In telescoping series, terms cancel out, leading to simpler partial sum expressions.
Convergence and Divergence of Series
Convergence and divergence are vital concepts in understanding series. Specifically, a series converges if its sequence of partial sums approaches a finite number. Conversely, a series diverges if the partial sums fail to settle at a specific value as the number of terms grows indefinitely.
For the series \( \sum_{n=1}^{\infty}(\tan(n) - \tan(n-1)) \), we find that the partial sum \( S_n = \tan(n) \). To determine the behavior of convergence, we examine the limit \( \lim_{n \to \infty} S_n \). Unfortunately, \( \tan(n) \), as the main term of our sum, does not stabilize to any finite value due to the periodic nature of the tangent function.
For the series \( \sum_{n=1}^{\infty}(\tan(n) - \tan(n-1)) \), we find that the partial sum \( S_n = \tan(n) \). To determine the behavior of convergence, we examine the limit \( \lim_{n \to \infty} S_n \). Unfortunately, \( \tan(n) \), as the main term of our sum, does not stabilize to any finite value due to the periodic nature of the tangent function.
- Convergence requires the partial sums to settle on a finite and consistent value.
- The tangent function's unbounded variations cause the series to diverge.
Infinite Series Analysis
Analyzing infinite series involves assessing the sum of infinitely many terms, a critical aspect of mathematics. Infinite series can portray functions or potential values, demanding a deep understanding of their behavior.
For the telescoping series \( \sum_{n=1}^{\infty}(\tan(n) - \tan(n-1)) \), we first identified its structure: the cancellation property typical of telescoping series. Applying this, we are left with the simplified expression for partial sums, \( S_n = \tan(n) \). By investigating whether this partial sum converges or diverges, we discovered divergence due to oscillations in \( \tan(n) \).
For the telescoping series \( \sum_{n=1}^{\infty}(\tan(n) - \tan(n-1)) \), we first identified its structure: the cancellation property typical of telescoping series. Applying this, we are left with the simplified expression for partial sums, \( S_n = \tan(n) \). By investigating whether this partial sum converges or diverges, we discovered divergence due to oscillations in \( \tan(n) \).
- Telescoping series simplify by canceling terms, showing only the significant ones.
- Understanding series behavior aids in determining convergence and potential value extraction.
Other exercises in this chapter
Problem 38
Determining Convergence or Divergence In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\
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Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 0} \frac{\sin 3 x^{2}}{1-\cos 2 x} \end{equation}
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How close is the approximation \(\sin x=x\) when \(|x|
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