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Which of the series in Exercises 1–36 converge, and which diverge? Give reasons for your answers. $$ \sum_{n=2}^{\infty} \frac{1}{n \sqrt{n^{2}-1}} $$

Short Answer

Expert verified
The series converges.

Step by step solution

01

Determine the form of the series

The given series is \( \sum_{n=2}^{\infty} \frac{1}{n \sqrt{n^{2}-1}} \). Notice that the series has terms of the form \( \frac{1}{n \sqrt{n^2 - 1}} \), which are quite complex but can be approximated or simplified for convergence testing.
02

Simplify the expression

Consider simplifying the term \( \sqrt{n^{2} - 1} \). For large values of \( n \), \( \sqrt{n^2-1} \approx n \). Therefore, \( \frac{1}{n \sqrt{n^{2}-1}} \approx \frac{1}{n^2} \) for large \( n \).
03

Choose the test for convergence

Since the simplified form resembles a \( p\)-series, we can use the \( p\)-series test. A \( p\)-series \( \sum \frac{1}{n^p} \) converges if \( p > 1 \). Here, \( p = 2 \) in the approximation \( \frac{1}{n^2} \).
04

Apply the Limit Comparison Test

To confirm, we use the Limit Comparison Test with the series \( \sum \frac{1}{n^2} \). Compute the limit:\[ L = \lim_{n \to \infty} \frac{\frac{1}{n \sqrt{n^{2}-1}}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n \sqrt{n^{2}-1}}. \]Simplify to find \( L \):\[ L = \lim_{n \to \infty} \frac{n}{\sqrt{n^{2}-1}} = \lim_{n \to \infty} \frac{n}{n\sqrt{1 - \frac{1}{n^2}}} = \lim_{n \to \infty} \frac{1}{\sqrt{1 - \frac{1}{n^2}}} = 1. \]Since \( L \) is a finite positive number, both series either converge or diverge together. \( \sum \frac{1}{n^2} \) is a convergent series, thus the original series converges.
05

Conclude the result

Since the Limit Comparison Test showed that the series \( \sum_{n=2}^{\infty} \frac{1}{n \sqrt{n^{2}-1}} \) converges by comparing it with the \( p\)-series \( \sum \frac{1}{n^2} \), we conclude that the given series converges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the P-Series Test
The p-series test is a fundamental tool in mathematical analysis when dealing with infinite series. It examines series of the form \( \sum \frac{1}{n^p} \). The test's core criterion is simple: if \( p > 1 \), the series converges; if \( p \leq 1 \), the series diverges. For instance, the harmonic series, where \( p = 1 \), diverges despite the decreasing size of its terms.
In this exercise, the series \( \sum \frac{1}{n \sqrt{n^2 - 1}} \) is simplified to approximate \( \frac{1}{n^2} \) for large \( n \), resembling a p-series with \( p = 2 \). Since \( 2 > 1 \), the simplified series converges, indicating that the original complex series likely does too. This test helps predict convergence through comparison to a well-understood form.
Exploring the Limit Comparison Test
The limit comparison test is another valuable tool for series convergence. It compares two series to determine their convergence behavior. To apply it, you calculate the limit of the ratio of their terms as \( n \) approaches infinity. Specifically, given series \( \sum a_n \) and \( \sum b_n \), the test asks us to find \[ L = \lim_{n \to \infty} \frac{a_n}{b_n} \]. If \( L \) is a finite positive number, both series either converge or diverge together.
In our example, the original series is compared with \( \sum \frac{1}{n^2} \), a known convergent p-series. Through simplification steps, it calculates \[ L = 1 \], solidifying the original series' convergence by matching it with the behavior of a converging p-series.
Grasping Series Convergence
Series convergence is about understanding whether the sum of an infinite number of terms settles at a certain value as you add more terms or keeps growing indefinitely. A series converges if, as you sum its terms, you approach a fixed number, the series doesn't blow up to infinity.
Convergence tests like the p-series test and the limit comparison test are handy here. By approximating complex series to simpler forms that we understand well, these tests help determine the convergence or divergence without having to sum an infinite number of terms. This makes complex problems manageable and provides deeper insight into their behavior.
The Role of Mathematical Analysis
Mathematical analysis involves the rigorous examination of methods like the p-series and limit comparison tests to study the behavior of series and sequences. It provides the framework for understanding convergence, divergence, and the conditions under which they occur.
Through mathematical analysis, we deconstruct a given series into a form that fits well-known tests, as demonstrated in this exercise. It allows us to handle seemingly complex forms using established logic, in-depth comparisons, and approximations, strengthening our problem-solving toolkit in calculus and beyond.
  • Analysis emphasizes clarity and precision in proofs and problem-solving.
  • It helps dissect complex problems into simpler parts using logical steps.
  • This rigorous approach ensures solutions are both correct and understandable
Enabling students to apply these tests, mathematical analysis enhances their capacity to tackle diverse challenges in infinite series.

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Most popular questions from this chapter

Sequences generated by Newton's method Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} $$ a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\) b. Starting with \(x_{0}=1\) and \(a=3\) , calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

Obtain the Taylor series for 1\(/(1+x)^{2}\) from the series for \(-1 /(1+x) .\)

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\sqrt[n]{n} $$

Compound interest, deposits, and withdrawals If you invest an amount of money \(A_{0}\) at a fixed annual interest rate \(r\) compounded \(m\) times per year, and if the constant amount \(b\) is added to the account at the end of each compounding period (or taken from the account if \(b<0 ),\) then the amount you have after \(n+1\) compounding periods is $$ A_{n+1}=\left(1+\frac{r}{m}\right) A_{n}+b $$ a. If \(A_{0}=1000, r=0.02015, m=12,\) and \(b=50\) , calculate and plot the first 100 points \(\left(n, A_{n}\right) .\) How much money is in your account at the end of 5 years? Does \(\left\\{A_{n}\right\\}\) converge? Is \(\left\\{A_{n}\right\\}\) bounded? b. Repeat part (a) with \(A_{0}=5000, r=0.0589, m=12,\) and \(b=-50 .\) c. If you invest 5000 dollars in a certificate of deposit (CD) that pays 4.5\(\%\) annually, compounded quarterly, and you make no further investments in the CD, approximately how many years will it take before you have \(20,000\) dollars? What if the CD earns 6.25\(\% ?\) d. It can be shown that for any \(k \geq 0\) , the sequence defined recursively by Equation \((1)\) satisfies the relation $$ A_{k}=\left(1+\frac{r}{m}\right)^{k}\left(A_{0}+\frac{m b}{r}\right)-\frac{m b}{r} $$ For the values of the constants \(A_{0}, r, m,\) and \(b\) given in part (a), validate this assertion by comparing the values of the first 50 terms of both sequences. Then show by direct substitution that the terms in Equation \((2)\) satisfy the recursion formula in Equation ( 1\()\) .

Estimate the error if \(\cos t^{2}\) is approximated by \(1-\frac{t^{4}}{2}+\frac{t^{8}}{4 !}\) in the integral \(\int_{0}^{1} \cos t^{2} d t .\)

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