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State what conclusion, if any, may be drawn from the Divergence Test. $$ \sum_{n=1}^{\infty} \frac{2 n^{2}}{3 n^{2}+n+1} $$

Short Answer

Expert verified
The series diverges by the Divergence Test.

Step by step solution

01

Understand the Divergence Test

The Divergence Test states that if \( \sum_{n=1}^{\infty} a_n \) is a series and \( \lim_{n \to \infty} a_n eq 0 \), then the series diverges. However, if \( \lim_{n \to \infty} a_n = 0 \), the test is inconclusive.
02

Identify the Series Terms

In this problem, the terms of the series are given by \( a_n = \frac{2n^2}{3n^2 + n + 1} \).
03

Calculate the Limit of the Series Terms

We calculate \( \lim_{n \to \infty} \frac{2n^2}{3n^2 + n + 1} \). To do this, divide every term by \( n^2 \), the highest power in the denominator: \[ \lim_{n \to \infty} \frac{2}{3 + \frac{1}{n} + \frac{1}{n^2}}. \]
04

Evaluate the Limit

As \( n \to \infty \), \( \frac{1}{n} \to 0 \) and \( \frac{1}{n^2} \to 0 \). Thus, the expression simplifies to \[ \lim_{n \to \infty} \frac{2}{3 + 0 + 0} = \frac{2}{3}. \]
05

Apply the Divergence Test

Since \( \lim_{n \to \infty} a_n = \frac{2}{3} eq 0 \), the Divergence Test concludes that the series \( \sum_{n=1}^{\infty} \frac{2 n^{2}}{3 n^{2}+n+1} \) diverges.

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Key Concepts

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

Series Divergence
When exploring infinite series in mathematics, one often encounters divergence. The Series Divergence, in particular, refers to a situation where the series does not converge to a finite sum. For instance, if we consider the series \( \sum_{n=1}^{\infty} \frac{2n^2}{3n^2 + n + 1} \), determining whether this series diverges or converges is crucial.
To assess divergence, we frequently use the Divergence Test. This test asserts that if the terms of the series do not approach zero as \( n \to \infty \), then the series diverges. In the example series, by evaluating \( \lim_{n \to \infty} a_n = \frac{2}{3} \), we see that the terms do not tend towards zero. Because of this, we can confidently declare that the series diverges.
Understanding divergence helps us avoid incorrect assumptions about the behavior of a series over infinite summation, emphasizing the importance of verifying conditions for convergence.
Convergence Tests
Convergence Tests are essential tools for determining whether an infinite series converges or diverges. Different tests apply to various types of series, offering math students a wealth of methods for tackling complex series.
One essential convergence test is the Divergence Test, which provides a simple yet powerful method to detect divergence. If the limit of the individual terms of a series does not equate to zero, the series will not converge. This approach was aptly illustrated in the given solution, where the series \( \sum_{n=1}^{\infty} \frac{2n^2}{3n^2 + n + 1} \) diverged as the term's limit was \( \frac{2}{3} \).
Beyond the Divergence Test, there are numerous other tests such as the Ratio Test, Root Test, and Integral Test. These tests evaluate conditions under which a series might converge, based on the behavior of its terms. Whenever you face a series problem, it's helpful to have these tests in your mathematical toolkit. They provide steps to verify whether your series behaves within set boundaries necessary for convergence.
Limit of a Sequence
The concept of the "Limit of a Sequence" is fundamental in understanding series. A sequence is a list of numbers in a particular order, and as we examine its behavior as \( n \to \infty \), we seek its limit if it exists.
If the terms in our sequence \( a_n \) approach a specific value, we say the limit is that value. We denote it as \( \lim_{n \to \infty} a_n \). This limit is vital when studying series, as it often determines the convergence or divergence of the overall series.
In the exercise provided, the sequence formed by \( a_n = \frac{2n^2}{3n^2 + n + 1} \) was analyzed for its limit. By simplifying the expression using division by the highest power, we found \( \lim_{n \to \infty} a_n = \frac{2}{3} \). Since the limit is not zero, the Divergence Test shows the series diverges.
The process of evaluating these limits teaches us not just about series and sequences but also about recognizing patterns and simplifying complex expressions for deeper insights into mathematical behavior.

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Most popular questions from this chapter

Consider the initial value problem $$ \frac{d y}{d x}=x^{2}+y, \quad y(0)=1 $$ \(\begin{array}{llll}\text { a. Calculate } & \text { the power } & \text { series } & \text { expansion }\end{array}\) \(y(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) of the solution up to the \(x^{7}\) term. b. Using the coefficients you have calculated, plot \(S_{3}(x)=\sum_{n=0}^{3} a_{n} x^{n}\) in the viewing rectangle \([-3,3] \times\) [-11,44] c. The exact solution to the initial value problem is \(y(x)=3 e^{x}-x^{2}-2 x-2,\) as can be determined using the methods of Section 7.7 in Chapter 7 . Add the plot of the exact solution to the viewing window. From the two plots, we see that the approximation is fairly accurate for \(-1 \leq x \leq 1\), but the accuracy decreases outside this subinterval. d. When a partial sum \(S_{N}(x)\) is used to approximate an infinite series, an increase in the value of \(N\) requires more computation, but improved accuracy is the reward. To see the effect in this example, replace the graph of \(S_{3}(x)\) with that of \(S_{7}(x)\).

Let \(f(x)=(x+1)^{4} /\left(x^{4}+1\right)\). It is known that \(f\) has a power series expansion of the form \(f(x)=1+4 x+6 x^{2}+4 x^{3}-4 x^{5}-6 x^{6}-4 x^{7}+4 x^{9}+\cdots\) a. Plot the central difference quotient approximation \(D_{0} f\left(x, 10^{-5}\right)\) of \(f^{\prime}(x)\) for \(-0.5

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. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{n}{n^{2}-11}\)

Use the Maclaurin series of \(\cos (x),\) namely \(1+u\) where \(u=\sum_{m=1}^{\infty}(-1)^{n} x^{2 n} /(2 n) !,\) and the Maclaurin series \(\sum_{m=1}^{\infty}(-1)^{m+1} u^{m} / m\) of \(\ln (1+u),\) to show that the Maclaurin series of \(\ln (\cos (x))\) is \(-x^{2} / 2-x^{4} / 12-x^{6} / 45-\) \(\ldots\) Use this result to obtain the Maclaurin series of tan \((x)\) up to and including the \(x^{5}\) term.

Prove that \(\sum_{n=1}^{\infty} \ln (1 / n)\) diverges.

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