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Magazine Chapter 8: Problem 54
Determine whether the following series converge absolutely, converge conditionally, or diverge. $$\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln k}$$
Short Answer
Expert verified
Answer: The series converges conditionally.
Step by step solution
01
Apply the Alternating Series Test
In order to apply the Alternating Series Test, we need to show that the terms of the sequence are decreasing and tending to zero. The sequence is given by:
$$a_k = \frac{(-1)^k}{\ln k}$$
Now, let's consider the sequence of positive values of \(a_k\), given by:
$$b_k = \frac{1}{\ln k}$$
As \(k\) increases, the natural logarithm, \(\ln k\), also increases. Consequently, the terms of the sequence \(b_k\) are decreasing. Now let's see if the limit of \(b_k\) as \(k\) approaches infinity is zero:
$$\lim_{k\to\infty} \frac{1}{\ln k} = 0$$
Since both conditions of the Alternating Series Test are satisfied, the series converges. We will now determine if it converges absolutely or conditionally.
02
Determine absolute convergence
To determine if the series converges absolutely, we will consider the absolute value of the series:
$$\sum_{k=2}^{\infty} \left|\frac{(-1)^{k}}{\ln k}\right| = \sum_{k=2}^{\infty} \frac{1}{\ln k}$$
Now we'll use the Comparison Test to check if this series converges or diverges.
03
Apply the Comparison Test
Let's compare this series with the integral of the function \(f(x) = \frac{1}{\ln x}\) from \(2\) to \(\infty\). We will do this by evaluating the integral:
$$\int_2^{\infty} \frac{1}{\ln x} dx$$
To evaluate this integral, we need to take a substitution \(u = \ln x\) and \(du = \frac{1}{x} dx\). The integral becomes:
$$\int \frac{1}{u} du = \ln |u| + C$$
Now we substitute back to get the result:
$$\ln|\ln x| + C$$
Now let's evaluate the limit of this expression as \(x \to \infty\):
$$\lim_{x\to\infty} \ln|\ln x| = \infty$$
Since the integral diverges, the comparison series also diverges. Therefore, the series converges, but it does not converge absolutely.
04
Conclusion
The given series \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln k}\) converges conditionally, as it converges by the Alternating Series Test but does not converge absolutely, according to the Comparison Test.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Alternating Series Test
When faced with an alternating series, a series where the terms alternate in sign, the Alternating Series Test comes into play as a useful tool for determining convergence. The test has two primary conditions that must be met for a series to be deemed convergent:
- The absolute value of the sequence's terms must be decreasing: $$a_{k+1} \text{.
- The limit of the sequence's terms must approach zero as the index goes to infinity: $$ lim_{k \to \text{is satisfied, the series is considered to converge. However, this doesn't automatically mean it converges absolutely鈥攊t could be conditional, which we'll explore in the following sections.
Absolute Convergence
Absolute convergence refers to a series where the sum of the absolute values of its terms converges. To determine absolute convergence, one evaluates the series without considering the signs of its terms.
- For the series $$ \text{If the series of absolute values diverges, then the series may still converge conditionally, depending on its alternating pattern, as per the Alternating Series Test.
Conditional Convergence
Conditional convergence occurs when a series converges, but it does not converge absolutely. This typically involves an alternating series where the negative and positive terms cancel each other out to some extent, allowing the series to converge even though the sum of the absolute values of the terms diverges.
- If a series passes the Alternating Series Test, but its corresponding absolute series does not converge, it is conditionally convergent鈥攁s seen in the example of $$\text{where the absolute series diverges (by the Comparison Test with an improper integral), thus rendering the original series conditionally convergent.
Comparison Test
The Comparison Test is a method used to determine the convergence or divergence of a series by comparing it to a known benchmark series or integral. If a series's terms can be proven to be less than or equal to another converging series's terms, then the original series also converges. Conversely, if the series terms are larger than that of a diverging series or integral, the original series diverges.
- For verifying absolute convergence, one might compare the series $$\text{. Since the integral diverges, we can infer that the series also diverges, only converging conditionally as proven earlier.
