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Chapter 7: Problem 72

State the \(n\) th-Term Test for Divergence.

Short Answer

Expert verified
The nth-Term Test for Divergence states that if the limit as \(n\) approaches infinity of the \(n\)th term of a series does not equal zero, the series is divergent.

Step by step solution

01

Understanding the nth-Term Test for Divergence

The nth-term test is a basic tool in calculus used to test the convergence or divergence of a series. It's straightforward and easy to apply. The big focus is on calculating the limit of the sequence forming the series.
02

Statement of the nth-Term Test for Divergence

The nth-Term Test for Divergence states: For a given series, if the limit as \(n\) approaches infinity of the \(n\)th term does not equal zero, then the series diverges. In mathematical terms, this is expressed as: If \(\lim_{n \to \infty} a_n \neq 0\), then the series \(\sum_{n=1}^\infty a_n\) is divergent.
03

Applications of the nth-Term Test for Divergence

It's important to note that if the \(n\)th term does go to zero, this test doesn't tell us anything about whether the series converges or diverges. For that, other series tests like the Integral Test, Comparison Test, or the Ratio Test might be utilized.

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Most popular questions from this chapter

(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0 . \overline{81} $$

(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0 . \overline{9} $$

(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0.0 \overline{75} $$

Let \(\left\\{x_{n}\right\\}, n \geq 0,\) be a sequence of nonzero real numbers such that \(x_{n}^{2}-x_{n-1} x_{n+1}=1\) for \(n=1,2,3, \ldots .\) Prove that there exists a real number \(a\) such that \(x_{n+1}=a x_{n}-x_{n-1},\) for all \(n \geq 1 .\)

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(|r|<1,\) then \(\sum_{n=1}^{\infty} a r^{n}=\frac{a}{(1-r)} .\)
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