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Chapter 8: Problem 4

For what values of \(p\) does the series \(\sum_{k=1}^{\infty} \frac{1}{k^{p}}\) converge? For what values of \(p\) does it diverge?

Short Answer

Expert verified
Answer: The series converges for p > 1 and diverges for p 鈮� 1.

Step by step solution

01

Express the series with the general formula

We are given the series \(\sum_{k=1}^{\infty} \frac{1}{k^{p}}\). Notice that the series starts at \(k = 1\) and goes to infinity.
02

Apply the Integral Test

Now, we will use the integral test to determine the convergence of the series. We need to compare the given series with the improper integral \(\int_{1}^{\infty} \frac{1}{x^{p}} dx\).
03

Evaluate the improper integral

First, we'll find the antiderivative of the function \(\frac{1}{x^{p}}\). The antiderivative is given by: $$ \int \frac{1}{x^{p}} dx = \frac{x^{1-p}}{1 - p} + C $$ where C is the constant of integration. Now, we'll evaluate the improper integral: $$ \int_{1}^{\infty} \frac{1}{x^{p}} dx = \lim_{b \to \infty} \left[ \frac{x^{1-p}}{1 - p} \right]_1^b = \lim_{b \to \infty} \frac{b^{1-p}}{1 - p} - \frac{1^{1-p}}{1 - p} $$
04

Determine the convergence of the improper integral

Now, we'll examine the limit of the improper integral as \(b\) goes to infinity: $$ \lim_{b \to \infty} \frac{b^{1-p}}{1 - p} - \frac{1^{1-p}}{1 - p} $$ We can now see that if \(p > 1\), the term \(b^{1-p}\) approaches 0 as \(b\) goes to infinity, and the integral converges. However, if \(p \le 1\), the term \(b^{1-p}\) either goes to infinity or stays constant, and the integral diverges.
05

Conclusion

Using the integral test, we find that the series converges for p > 1 and diverges for p 鈮� 1. Therefore, the series \(\sum_{k=1}^{\infty} \frac{1}{k^{p}}\) converges for \(p > 1\) and diverges for \(p \le 1\).

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Most popular questions from this chapter

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\int_{1}^{n} x^{-2} d x$$

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{1}{n^{2}}=0$$

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} 9(0.1)^{k}$$

In Section \(8.3,\) we established that the geometric series \(\sum r^{k}\) converges provided \(|r| < 1\). Notice that if \(-1 < r<0,\) the geometric series is also an alternating series. Use the Alternating Series Test to show that for \(-1 < r <0\), the series \(\sum r^{k}\) converges.

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)
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