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Chapter 7: Problem 123

Show that the Root Test is inconclusive for the \(p\) -series \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\)

Short Answer

Expert verified
The Root Test is inconclusive for the p-series \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\)

Step by step solution

01

Evaluate nth term

First, evaluate the nth term of the series which is \(a_{n} = \frac{1}{n^{p}}\).
02

Apply the Root Test

Now, apply the Root Test to the nth term. This involves taking the nth root of the absolute value of the nth term, which is \( \left(\frac{1}{n^{p}} \right)^{\frac{1}{n}} \).
03

Simplify the expression

Simplify the expression \( \left(\frac{1}{n^{p}} \right)^{\frac{1}{n}} \) to \( n^{-p/n} \).
04

Take the limit as n goes to inifnity

Evaluate the limit of \( n^{-p/n} \) as n tends to infinity. The limit equals 1.
05

Conclusion

Since the limit equals 1, the root test is inconclusive. It doesn't provide proof for convergence or divergence of the series.

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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 $$ 0 . \overline{9} $$

(a) find the common ratio of the geometric series, \((b)\) write the function that gives the sum of the series, and (c) use a graphing utility to graph the function and the partial sums \(S_{3}\) and \(S_{5} .\) What do you notice? $$ 1-\frac{x}{2}+\frac{x^{2}}{4}-\frac{x^{3}}{8}+\cdots $$

Give an example of a sequence satisfying the condition or explain why no such sequence exists. (Examples are not unique.) A sequence that converges to \(\frac{3}{4}\)

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

Describe the difference between \(\lim _{n \rightarrow \infty} a_{n}=5\) and \(\sum_{n=1}^{\infty} a_{n}=5\).
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