Problem 33
Question
Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=2}^{\infty} \frac{1}{n \sqrt{n^{2}-1}} $$
Step-by-Step Solution
Verified Answer
The series converges by the Limit Comparison Test with \( \frac{1}{n^2} \).
1Step 1: Analyze the Expression
The series given is \( \sum_{n=2}^{\infty} \frac{1}{n \sqrt{n^{2}-1}} \). To determine whether it converges or diverges, we need to consider the expression under the summation and compare it with a known series.
2Step 2: Simplify the Denominator
Rewrite the expression \( \sqrt{n^2 - 1} \) using an approximation for large \( n \); it behaves like \( \sqrt{n^2} = n \). Thus, the term becomes \( \frac{1}{n \cdot n} = \frac{1}{n^2} \) when \( n \) is large.
3Step 3: Choose a Comparison Series
The series \( \sum \frac{1}{n^2} \) is a known convergent p-series, with \( p = 2 > 1 \). This will be used to compare with the given series.
4Step 4: Apply the Limit Comparison Test
Using the Limit Comparison Test, consider \( a_n = \frac{1}{n \sqrt{n^2 - 1}} \) and \( b_n = \frac{1}{n^2} \). Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{n \sqrt{n^2 - 1}}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n \sqrt{n^2 - 1}} \).
5Step 5: Evaluate the Limit
Simplify the expression: \( \lim_{n \to \infty} \frac{n^2}{n \cdot \sqrt{n^2 - 1}} = \lim_{n \to \infty} \frac{n^2}{n \cdot n} \approx \lim_{n \to \infty} \frac{n^2}{n^2} = 1 \).
6Step 6: Conclude Using the Limit Comparison Test
Since the limit equals 1, and \( \sum \frac{1}{n^2} \) converges, then by the Limit Comparison Test, \( \sum \frac{1}{n \sqrt{n^2 - 1}} \) also converges.
Key Concepts
Limit Comparison TestConvergent p-seriesSeries Approximation
Limit Comparison Test
The Limit Comparison Test is a powerful tool for determining the convergence or divergence of a series. It is particularly useful when dealing with series that are not immediately recognizable as belonging to a specific category of simple series, like geometric or p-series.
The idea is simple:- We compare the series in question with a "comparison series" that has known behavior (either convergent or divergent).
- We compute the limit of the ratio of the terms of the two series, \[\lim_{n \to \infty} \frac{a_n}{b_n},\]where \( a_n \) is the nth term of the series you are investigating and \( b_n \) is the nth term of the known series.
If the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \) exists and is a finite number greater than zero, then the two series either both converge or both diverge. This makes it a very convenient method to apply as long as you can find an appropriate comparison series.
For example, with the series in the exercise, we compared it to the p-series \( \sum \frac{1}{n^2} \), a known convergent series.
The idea is simple:- We compare the series in question with a "comparison series" that has known behavior (either convergent or divergent).
- We compute the limit of the ratio of the terms of the two series, \[\lim_{n \to \infty} \frac{a_n}{b_n},\]where \( a_n \) is the nth term of the series you are investigating and \( b_n \) is the nth term of the known series.
If the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \) exists and is a finite number greater than zero, then the two series either both converge or both diverge. This makes it a very convenient method to apply as long as you can find an appropriate comparison series.
For example, with the series in the exercise, we compared it to the p-series \( \sum \frac{1}{n^2} \), a known convergent series.
Convergent p-series
Understanding p-series is crucial because they provide a benchmark for determining the behavior of other series through tests like the Limit Comparison Test.
A p-series is of the form:\[\sum_{n=1}^{\infty} \frac{1}{n^p}\]where \( p \) is a positive constant. The convergence of a p-series depends on the value of \( p \):- If \( p > 1 \), the series converges.- If \( p \leq 1 \), the series diverges.
In the given exercise, the series \( \sum \frac{1}{n^2} \) is used as a comparison. This is a classic example of a p-series with \( p = 2 \), which is known to converge. By recognizing similar terms or approximations to the series in question during the limit comparison, we can effectively predict if the series will converge or not.
This property of p-series allows them to act as a reference point, helping to simplify or rearrange other series to gain insights about their behavior.
A p-series is of the form:\[\sum_{n=1}^{\infty} \frac{1}{n^p}\]where \( p \) is a positive constant. The convergence of a p-series depends on the value of \( p \):- If \( p > 1 \), the series converges.- If \( p \leq 1 \), the series diverges.
In the given exercise, the series \( \sum \frac{1}{n^2} \) is used as a comparison. This is a classic example of a p-series with \( p = 2 \), which is known to converge. By recognizing similar terms or approximations to the series in question during the limit comparison, we can effectively predict if the series will converge or not.
This property of p-series allows them to act as a reference point, helping to simplify or rearrange other series to gain insights about their behavior.
Series Approximation
Approximating complex mathematical expressions in series can simplify calculations, especially as the number of terms grows large.
In the given problem, the term \( \sqrt{n^2 - 1} \) appears in the denominator, making it cumbersome for comparison. However, when \( n \) is large, mathematical approximations help to simplify \( \sqrt{n^2 - 1} \) to something more manageable. As \( n \to \infty \), the expression \( \sqrt{n^2 - 1} \) approximates closely to \( n \), because the effect of the "-1" becomes negligible relative to \( n^2 \). Thus, the term simplifies to \( \frac{1}{n^2} \) for large \( n \).
This approximation process is crucial when applying comparison tests, as it allows you to better align your series with a known comparison series. By understanding and applying approximations correctly, the process of determining convergence through series analysis becomes much more straightforward.
This approach not only simplifies complex expressions but also heightens your ability to connect seemingly complex series to simpler ones, such as the classic p-series, facilitating the entire convergence assessment.
In the given problem, the term \( \sqrt{n^2 - 1} \) appears in the denominator, making it cumbersome for comparison. However, when \( n \) is large, mathematical approximations help to simplify \( \sqrt{n^2 - 1} \) to something more manageable. As \( n \to \infty \), the expression \( \sqrt{n^2 - 1} \) approximates closely to \( n \), because the effect of the "-1" becomes negligible relative to \( n^2 \). Thus, the term simplifies to \( \frac{1}{n^2} \) for large \( n \).
This approximation process is crucial when applying comparison tests, as it allows you to better align your series with a known comparison series. By understanding and applying approximations correctly, the process of determining convergence through series analysis becomes much more straightforward.
This approach not only simplifies complex expressions but also heightens your ability to connect seemingly complex series to simpler ones, such as the classic p-series, facilitating the entire convergence assessment.
Other exercises in this chapter
Problem 33
Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=\cos (2 x+(\pi / 2)), \quad a=\pi / 4\)
View solution Problem 33
Use series to evaluate the limits. \begin{equation} \lim _{y \rightarrow 0} \frac{y-\tan ^{-1} y}{y^{3}} \end{equation}
View solution Problem 33
In Exercises \(1-36,\) (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely (c) conditional
View solution Problem 33
In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{
View solution