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Chapter 9: Problem 94

State the Root Test.

Short Answer

Expert verified
The Root Test involves finding the limit of the nth root of the absolute value of the nth term of a series as n approaches infinity. If the limit is less than 1, the series converges. If it's more than 1, the series diverges. If the limit is 1, the test is inconclusive.

Step by step solution

01

Understanding the Root Test

The Root Test is used to determine whether a series converges or diverges. If we have a series \(\sum a_n\), we can take the nth root of the absolute value of each term, \(|a_n|^{1/n}\), then determine the limit as n approaches infinity.
02

Determining the convergence or divergence based on the limit

If the limit \(L = \lim_{n \to \infty} |a_n|^{1/n}\) is less than 1, the series converges. If the limit \(L\) is greater than 1, the series diverges. If \(L = 1\), the test is inconclusive, which means we cannot be sure whether the series converges or diverges based on this test.
03

Using the Root Test in Practice

In practice, the root test is most commonly used when each term of the series is raised to the power of n, because taking the nth root will simplify the term. Apply the root test by first determining the limit and then comparing it with 1 to decide the convergence or divergence of the series.

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Most popular questions from this chapter

Finding a Taylor Series In Exercises \(1-12,\) use the definition of Taylor series to find the Taylor series, centered at \(c,\) for the function. $$ f(x)=\ln x, \quad c=1 $$

Finding a Maclaurin Series In Exercises \(41-44\) , find the Maclaurin series for the function. (See Examples 7 and \(8 . )\) $$ h(x)=x \cos x $$

Approximating an Integral In Exercises \(63-70\) , use a power series to approximate the value of the integral with an error of less than \(0.0001 .\) (In Exercises 65 and \(67,\) assume that the integrand is defined as 1 when \(x=0 .\) $$ \int_{0}^{0.2} \sqrt{1+x^{2}} d x $$

Verifying a Formula In Exercises 45 and \(46,\) use a power series and the fact that \(i^{2}=-1\) to verify the formula. $$ g(x)=\frac{1}{2 i}\left(e^{i x}-e^{-i x}\right)=\sin x $$

Is the following series convergent or divergent? \(1+\frac{1}{2} \cdot \frac{19}{7}+\frac{2 !}{3^{2}}\left(\frac{19}{7}\right)^{2}+\frac{3 !}{4^{3}}\left(\frac{19}{7}\right)^{3}+\frac{4 !}{5^{4}}\left(\frac{19}{7}\right)^{4}+\cdots\)
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