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Chapter 10: Problem 34

Use the Divergence Test, the Integral Test, or the p-series test to determine whether the following series converge. $$\sum_{k=3}^{\infty} \frac{1}{k(\ln k) \ln \ln k}$$

Short Answer

Expert verified
Answer: The series diverges.

Step by step solution

01

Determine type of series

The given series is: $$\sum_{k=3}^{\infty} \frac{1}{k(\ln k) \ln (\ln k)}$$ The series doesn't look like a \(p\)-series, nor does the limit of its terms go to \(0\) as \(k\) goes to infinity. So, we should use the Integral Test.
02

Apply the Integral Test

The Integral Test states that if the function \(f(x)\) is continuous, positive, and decreasing for \(x \ge k\), then our summation, \(\sum_{k=3}^{\infty} f(k)\), converges if and only if the integral \(\int_{k}^{\infty} f(x)dx\) converges. Our function \(f(k) = \frac{1}{k(\ln k) \ln (\ln k)}\) is positive, continuous, and decreasing for \(k \ge 3\), so we can apply the Integral Test.
03

Evaluate the integral

We now have to evaluate the improper integral: $$\int_{3}^{\infty} \frac{1}{x(\ln x) \ln (\ln x)}dx$$ To do this, we can use the substitution technique. Let \(u =\ln x\). Then, \(du =\frac{1}{x}dx\). So, we have: $$\int_{\ln 3}^{\infty} \frac{du}{u \ln u}$$ Now we can make one more substitution. Let \(v = \ln u\). Then, \(dv =\frac{1}{u}du\). So, we get: $$\int_{\ln(\ln 3)}^{\infty} \frac{dv}{v}$$ Now, we have to evaluate this integral: $$\lim_{t \to \infty} \int_{\ln(\ln 3)}^{t} \frac{dv}{v} = \lim_{t \to \infty} (\ln v) \Big|_{\ln(\ln 3)}^{t} = \lim_{t \to \infty} (\ln t - \ln(\ln (\ln 3)))$$ The limit as \(t\) goes to infinity is infinity, which means that the integral diverges.
04

Conclude using the Integral Test

Since the integral diverges, the series also diverges according to the Integral Test. Therefore, the given series: $$\sum_{k=3}^{\infty} \frac{1}{k(\ln k) \ln \ln k}$$ diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divergence Test
The Divergence Test, also known as the Test for Divergence, is a quick way to determine if a series does not converge. It's one of the simplest tests to apply. If the terms of a series do not approach zero as the number of terms goes to infinity, the series must diverge. For example, consider the series
  • \(\sum_{k} a_k\).
If \(\lim_{{k \to \infty}} a_k eq 0\), then the series diverges. However, if the limit of the terms of the series does go to zero, you can't conclude convergence; you need to test further. Therefore, while this test is useful, it primarily confirms divergence, not convergence. Often, other tests are required to establish convergence.
Integral Test
The Integral Test is a valuable tool for determining the convergence of a series, especially when the terms are derived from functions that are continuous, positive, and decreasing. The idea is to compare the given series
  • \( \sum_{k = a}^{\infty} f(k) \)
with an improper integral
  • \( \int_{a}^{\infty} f(x)dx \).
If the integral converges, the series converges. If the integral diverges, the series diverges as well. For application, ensure that the function \( f(x) \) maintains its properties of being continuous, positive, and decreasing over the interval. The step-by-step substitution method can help in evaluating complex integrals, as demonstrated in evaluating
  • \(\int_{3}^{\infty} \frac{1}{x(\ln x) \ln (\ln x)}dx \).
Using substitutions, the integral was simplified and evaluated, leading to the conclusion that the series diverges.
p-series
A p-series is a series of the form
  • \( \sum_{k=1}^{\infty} \frac{1}{k^p} \)
where \( p \) is a positive constant. The convergence or divergence of a p-series depends solely on the value of \( p \):
  • If \( p > 1 \), the series converges.
  • If \( p \leq 1 \), the series diverges.
This makes p-series straightforward to analyze. In our example problem, the series given was not a p-series, as it involved logarithmic terms in the denominator. Therefore, the p-series test was not suitable for this scenario, prompting the use of the Integral Test instead.

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Most popular questions from this chapter

Express each sequence \(\left\\{a_{n}\right\\}_{n=1}^{\infty}\) as an equivalent sequence of the form \(\left\\{b_{n}\right\\}_{n=3}^{\infty}\). $$\left\\{n^{2}+6 n-9\right\\}_{n=1}^{\infty}$$

Determine whether the following series converge. Justify your answers. $$\sum_{k=1}^{\infty} \frac{7 k^{2}-k-2}{4 k^{4}-3 k+1}$$

Determine whether the following series converge. Justify your answers. $$\sum_{k=0}^{\infty} k \cdot 3^{-k^{2}}$$

The expression $$\begin{aligned} &1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+}}}}\\\ &\end{aligned}$$ where the process continues indefinitely, is called a continued fraction. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=1+\frac{1}{a_{n}}\) for \(n=0,1,2,3, \ldots\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n},\) provided the limit exists. b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\) c. Using computation and/or graphing, estimate the limit of the sequence. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. Compare your estimate with \(\frac{1+\sqrt{5}}{2},\) a number known as the golden mean. e. Assuming the limit exists, use the same ideas to determine the value of $$\begin{aligned} &a+\frac{b}{a+\frac{b}{a+\frac{b}{a+\frac{b}{a+}}}}\\\ &\end{aligned}$$ where \(a\) and \(b\) are positive real numbers.

Identify a convergence test for each of the following series. If necessary, explain how to simplify or rewrite the series before applying the convergence test. You do not need to carry out the convergence test. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{\sqrt{2^{k}+\ln k}}$$
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