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Use a Comparison Test to determine whether the given series converges or diverges. $$ \sum_{n=1}^{\infty}\left(\sec \left(\frac{1}{\sqrt{n}}\right)-\cos \left(\frac{1}{\sqrt{n}}\right)\right) $$

Short Answer

Expert verified
The series diverges because it behaves like the harmonic series \( \sum \frac{1}{n} \).

Step by step solution

01

Identify the Series Components

The given series is \( \sum_{n=1}^{\infty}\left(\sec \left(\frac{1}{\sqrt{n}}\right)-\cos \left(\frac{1}{\sqrt{n}}\right)\right) \). The terms involve the functions \( \sec(x) = \frac{1}{\cos(x)} \) and \( \cos(x) \). Our task is to find a suitable comparison series to determine convergence.
02

Analyze Behavior of Terms

As \( n \to \infty \), \( \frac{1}{\sqrt{n}} \to 0 \). For small values of \( x \), \( \sec(x) \approx 1 + \frac{x^2}{2} \) and \( \cos(x) \approx 1 - \frac{x^2}{2} \). Thus, for large \( n \), the term \( \sec\left(\frac{1}{\sqrt{n}}\right) - \cos\left(\frac{1}{\sqrt{n}}\right) \approx \frac{\left(\frac{1}{\sqrt{n}}\right)^2}{1} = \frac{1}{n} \).
03

Define the Comparison Series

The behavior of \( a_n = \sec\left(\frac{1}{\sqrt{n}}\right) - \cos\left(\frac{1}{\sqrt{n}}\right) \approx \frac{1}{n} \) suggests using the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) for comparison. This is a well-known divergent series.
04

Apply the Limit Comparison Test

Compute \( \lim_{n \to \infty} \frac{\sec\left(\frac{1}{\sqrt{n}}\right) - \cos\left(\frac{1}{\sqrt{n}}\right)}{\frac{1}{n}} \). From earlier approximation, this limit simplifies to \( 1 \), indicating that the behavior of the series \( a_n \) is similar to \( \frac{1}{n} \).
05

Conclusion

Since \( \sum_{n=1}^{\infty} \frac{1}{n} \) is a divergent series and the limit comparison test shows that \( \sum_{n=1}^{\infty} \left( \sec\left(\frac{1}{\sqrt{n}}\right) - \cos\left(\frac{1}{\sqrt{n}}\right) \right) \) behaves similarly, we conclude that the original series also diverges.

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Key Concepts

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

Series Convergence
When we talk about series convergence, we are discussing whether the sum of an infinite sequence of numbers approaches a certain value or not. If the series does approach a finite number, it converges; if not, it diverges.
A simple way to picture this is by imagining you are adding up numbers forever. If your total keeps getting closer and closer to a particular number (like 5, 100, etc.), then your series is converging to that number. If the total grows indefinitely or does not settle around a number, it diverges.
  • **Convergent Series**: The series' total approaches a specific value.
  • **Divergent Series**: The series' total does not approach any particular value; it either grows to infinity or doesn't settle down.
Understanding whether a series converges or diverges is crucial for many applications in math and science.
It can define behaviors and outcomes in calculus, analysis, and beyond.
Limit Comparison Test
The Limit Comparison Test is a handy tool for figuring out if a series converges or diverges. It allows us to compare another known series to our series of interest. If both series behave similarly in terms of convergence, then we can draw conclusions. Here's how it works:
Imagine you have a series, say \( a_n \), and you want to check if it converges or diverges. You find another series \( b_n \), which you already know converges or diverges.
  • You calculate the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
  • If this limit is a positive finite number, it means \( a_n \) behaves similarly to \( b_n \). So, if \( b_n \) converges, \( a_n \) converges, and vice versa.
For example, in our exercise, we estimated that \( a_n = \sec\left(\frac{1}{\sqrt{n}}\right) - \cos\left(\frac{1}{\sqrt{n}}\right) \approx \frac{1}{n} \). The series \( \sum \frac{1}{n} \) is known to diverge, so through our Limit Comparison Test, we conclude that our series also diverges. This test can drastically simplify the process of understanding series behavior.
Divergent Series
Divergent series are like a never-ending journey with no destination in sight; the sum keeps increasing without settling.
Understanding divergent series is important because it tells us the limitations of growth and accumulation in mathematical terms.
  • **Harmonic Series:** One famous divergent series is the harmonic series: \( \sum_{n=1}^{\infty} \frac{1}{n} \). The term \( \frac{1}{n} \) gets smaller as \( n \) increases, yet the series itself doesn't settle, it keeps growing.
  • **Implications**: Divergence indicates that the series doesn't "add up" to a finite value, often implying something is infinitely large.
In our exercise, comparing to the harmonic series, our original series \( \sum_{n=1}^{\infty} \left( \sec\left(\frac{1}{\sqrt{n}}\right) - \cos\left(\frac{1}{\sqrt{n}}\right) \right) \) was shown to be divergent. This tells us the sum grows indefinitely much like the harmonic series. Thus, understanding divergence is essential to correctly interpret mathematical phenomena and avoid incorrect conclusions about infinite sum behavior.

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Most popular questions from this chapter

Determine whether the series converges absolutely, converges conditionally, or diverges. The tests that have been developed in Section 5 are not the most appropriate for some of these series. You may use any test that has been discussed in this chapter. \(\sum_{n=2}^{\infty}(-1)^{n} \frac{1}{n \ln ^{3}(n)}\)

If \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\) converge on \((-R, R),\) then we may formally multiply the series as though they were polynomials. That is, if \(h(x)=f(x) g(x),\) then $$ h(x)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n} a_{k} b_{n-k}\right) x^{n} $$ The product series, which is called the Cauchy product, also converges on \((-R, R) .\) Exercises \(59-62\) concern the Cauchy product. Suppose \(|x|<1 .\) Calculate the power series of \(h(x)=1 /\) \(\left(1-x^{2}\right)\) with base point 0 by substituting \(t=x^{2}\) the equation \(1 /(1-t)=\sum_{n=0}^{\infty} t^{n} .\) Let \(f(x)=1 /(1-x)\) and \(g(x)=f(-x)\). Verify the Cauchy product formula for \(h=f \cdot g\) up to the \(x^{8}\) term.

Use a Taylor polynomial with base point \(c=e^{3}\) to approximate \(\ln (20)\) to five decimal places. Do not use a calculator to evaluate any value of \(\ln (x),\) but you may use a calculator for arithmetic with the number \(e\) and its powers.

Let \(\left\\{a_{n}\right\\}\) be a sequence of positive numbers. In a course on mathematical analysis, one learns that if the two limits \(\lim _{n \rightarrow \infty} a_{n+1} / a_{n}\) and \(\lim _{n \rightarrow \infty} a_{n}^{1 / n}\) exist, then they are equal. In each of Exercises \(65-68\), produce a plot that illustrates the equality of these two limits. Your plot should include a horizontal line that is the asymptote of the points \(\left\\{\left(n, a_{n+1} / a_{n}\right)\right\\}\) and \(\left\\{\left(n, a_{n}^{1 / n}\right)\right\\}\). \(a_{n}=e^{n} / \ln (1+n)\)

Determine whether the series converges absolutely, converges conditionally, or diverges. The tests that have been developed in Section 5 are not the most appropriate for some of these series. You may use any test that has been discussed in this chapter. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{\ln (n)}{\sqrt{n}}\)

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