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Chapter 8: Problem 70

Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{(\ln k)^{p}}$$

Short Answer

Expert verified
Answer: The series converges for \(p > 1\).

Step by step solution

01

Determine the function f(x)

Begin by identifying the function \(f(x)\) that corresponds to the given series. In this case, we have: $$f(x) = \frac{1}{(\ln x)^{p}}$$
02

Check the conditions for the integral test

To apply the integral test, we need to ensure that \(f(x)\) is positive, continuous, and decreasing on the interval \([2, \infty)\). Since \(\ln x > 0\) for \(x > 1\), and p > 0, it follows that \(f(x)\) is positive on the interval \((1, \infty)\). As \(f(x)\) is a composite of the natural logarithm function and a power function, it is also continuous on the interval \((1, \infty)\). Finally, to confirm that \(f(x)\) is decreasing, we can take the derivative: $$f'(x) = -\frac{p}{x (\ln x)^{p + 1}}$$ Since \(x > 1\), \(p > 0\), and \(\ln x > 0\), we have \(f'(x) < 0\) on the interval \((1, \infty)\). This means that \(f(x)\) is decreasing on the interval \((1, \infty)\), and we can apply the integral test.
03

Calculate the integral

Now, we will compute the integral: $$\int_{2}^{\infty} \frac{1}{(\ln x)^{p}} dx$$ Make the substitution \(u = \ln x\) and \(du = \frac{1}{x} dx\), we have: $$\int_{\ln2}^{\infty} \frac{1}{u^{p}} du$$ Now we use the power rule for integration: $$\int u^{-p} du = \frac{u^{-p+1}}{-p+1} + C$$ Applying the limits, we have: $$\lim_{T\to\infty} \left[ \frac{u^{-p+1}}{-p+1} \right]_{\ln 2}^{T}$$ When \(p > 1\), the integral converges: $$\frac{1}{1-p}\left(\lim_{T\to\infty} {T^{-p+1}}-\left(\ln2\right)^{1-p}\right)=\frac{1}{1-p}\left(0-\left(\ln2\right)^{1-p}\right)$$ When \(p \le 1\), the integral diverges: $$\lim_{T\to\infty} \left[ \frac{u^{-p+1}}{-p+1} \right]_{\ln 2}^{T} = \infty$$
04

Determine the values of p for which the series converges

Since the integral test tells us that the series and the integral have the same convergence properties (either both converge or both diverge), we can use the results from Step 3 to determine the values of p for which the series converges. From our calculations, the series converges when \(p > 1\) and diverges otherwise. Therefore, the values of the parameter p for which the given series converges are \(p > 1\).

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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. $$a_{n}=\tan ^{-1}\left(\frac{10 n}{10 n+4}\right)$$

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=\sqrt{2+a_{n}} ; a_{0}=1$$

For a positive real number \(p,\) the tower of exponents \(p^{p^{p}}\) continues indefinitely and the expression is ambiguous. The tower could be built from the top as the limit of the sequence \(\left\\{p^{p},\left(p^{p}\right)^{p},\left(\left(p^{p}\right)^{p}\right)^{p}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{1}=p^{p} .\) The tower could also be built from the bottom as the limit of the sequence \(\left\\{p^{p}, p^{\left(p^{p}\right)}, p^{\left(p^{(i)}\right)}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=p^{a_{n}}(\text { building from the bottom })\) where again \(a_{1}=p^{p}\). a. Estimate the value of the tower with \(p=0.5\) by building from the top. That is, use tables to estimate the limit of the sequence defined recursively by (1) with \(p=0.5 .\) Estimate the maximum value of \(p > 0\) for which the sequence has a limit. b. Estimate the value of the tower with \(p=1.2\) by building from the bottom. That is, use tables to estimate the limit of the sequence defined recursively by (2) with \(p=1.2 .\) Estimate the maximum value of \(p > 1\) for which the sequence has a limit.

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Evaluate the series \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)\) two ways. a. Use a telescoping series argument. b. Use a geometric series argument after first simplifying \(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\)
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