/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Test the following series for co... [FREE SOLUTION] | 91Ó°ÊÓ

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Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} a_{n} \quad \text { if } a_{n+1}=\frac{n}{2 n+3} a_{n}$$

Short Answer

Expert verified
The series \( \sum_{n=1}^{\infty} a_{n} \) converges by the Ratio Test.

Step by step solution

01

- Identify the Series

We are given the series \( \sum_{n=1}^{\infty} a_{n} \) with the recurrence relation \( a_{n+1} = \frac{n}{2n+3} a_{n} \).
02

- Apply the Ratio Test

For the Ratio Test, consider the limit \( L = \lim_{n \to \infty}\frac{a_{n+1}}{a_{n}} \). Given \( a_{n+1} = \frac{n}{2n+3} a_{n} \), we have \( L = \lim_{n \to \infty} \frac{n}{2n+3} \).
03

- Simplify the Limit

Simplify the limit by dividing the numerator and the denominator by \( n \). This gives \( L = \lim_{n \to \infty} \frac{1}{2 + \frac{3}{n}} \). As \( n \to \infty \), the term \( \frac{3}{n} \to 0, \) so \( L = \frac{1}{2} \).
04

- Interpret the Ratio Test Result

Since \( L = \frac{1}{2} < 1 \), the Ratio Test indicates that the series \( \sum_{n=1}^{\infty} a_{n} \) converges.

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Key Concepts

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

ratio test
The Ratio Test is a powerful tool for determining the convergence or divergence of an infinite series. To use the Ratio Test, you calculate the limit of the absolute value of the ratio of consecutive terms. In mathematical terms, if you have a series given by \( \sum_{n=1}^{\infty} a_{n} \), you consider the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| \).
If \( L < 1 \), the series converges. If \( L > 1 \), it diverges. When \( L = 1 \), the test is inconclusive.

In our given exercise, the Ratio Test helped us find that \( L = \frac{1}{2} \), indicating convergence since \( \frac{1}{2} \) is less than 1.

This is a straightforward way to determine the behavior of an infinite series, making the Ratio Test a convenient choice for series involving complex expressions.
recurrence relation
A recurrence relation expresses each term of a series based on the previous terms. In the context of the series \( \sum_{n=1}^{\infty} a_{n} \), the recurrence relation given is \( a_{n+1} = \frac{n}{2n+3} a_{n} \).

This means that to find any term in the series, you multiply the preceding term by \( \frac{n}{2n+3} \).
Recurrence relations are helpful in breaking down complex series into simpler components, making it easier to analyze their behavior.

By using the recurrence relation, we can apply tests like the Ratio Test, as it gives us a clear way to compare successive terms.
series convergence
Series convergence refers to whether the sum of the terms in a series approaches a finite value as the number of terms goes to infinity. There are various tests to check for series convergence, including the Ratio Test, which was employed in our exercise.

When the limit of the ratio of consecutive terms is less than one, the series converges. In our case, we found that \( L = \frac{1}{2} \), which clearly shows that the series converges.

Understanding the concept of convergence is crucial as it allows us to determine whether the infinite sum will result in a finite number.
infinite series
An infinite series is a sum of infinitely many terms. The notation \( \sum_{n=1}^{\infty} a_{n} \) expresses the idea of adding up an endless sequence of terms given by \( a_{n} \).

Studying infinite series is vital in various fields of mathematics and applied sciences. It helps in analyzing patterns and behaviors that recur indefinitely.

When dealing with infinite series, determining whether they converge or diverge is key. Convergent series have a finite sum, while divergent series do not. Applying convergence tests, such as the Ratio Test, helps us make this determination. In our exercise, we used these principles to find that the given series converges.

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Most popular questions from this chapter

(a) Using computer or tables (or see Chapter 7 , Section 11 ), verify that \(\sum_{n=1}^{\infty} 1 / n^{2}=\) \(\pi^{2} / 6=1.6449+,\) and also verify that the error in approximating the sum of the series by the first five terms is approximately 0.1813 . (b) \(\quad\) By computer or tables verify that \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)(1 / 2)^{n}=\pi^{2} / 12-(1 / 2)(\ln 2)^{2}=\) \(0.5822+,\) and that the sum of the first five terms is \(0.5815+.\) (c) Prove theorem (14.4). Hint: The error is \(\left|\sum_{N+1}^{\infty} a_{n} x^{n}\right| .\) Use the fact that the absolute value of a sum is less than or equal to the sum of the absolute values. Then use the fact that \(\left|a_{n+1}\right| \leq\left|a_{n}\right|\) to replace all \(a_{n}\) by \(a_{N+1},\) and write the appropriate inequality. Sum the geometric series to get the result.

Use series you know to show that: $$\ln 3+\frac{(\ln 3)^{2}}{2 !}+\frac{(\ln 3)^{3}}{3 !}+\cdots=2$$

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$$

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\ln \left(2-e^{-x}\right)$$

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{\sin x-x}{x^{3}}$$

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