Problem 18 - Single Variable Calculus

Question

Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent. \(\sum_{n=2}^{\infty} \frac{(-1)^{n+1} \ln n}{n^{2}+1}\)

Step-by-Step Solution

Answer

The original series, \(∑_{n=2}^{\infty} \frac{(-1)^{n+1} \ln n}{n^{2}+1}\), is absolutely convergent.

Step 1: Find the absolute value of the series

To find the absolute value of the series, we need to remove the alternating sign by taking the absolute value of each term. The absolute value of the given series is: \[|\sum_{n=2}^{\infty} \frac{(-1)^{n+1} \ln n}{n^{2}+1}| = ∑_{n=2}^{\infty} \frac{\ln n}{n^{2}+1}\] Step 2: Apply the Comparison Test

Step 2: Comparison Test

Since we have the absolute value of the series, we can use the Comparison Test to check for convergence. Remember that for the Comparison Test, we are looking for a similar series that is easier to determine convergence. Notice that the series inside the summation is smaller than or equal to the series where the denominator is only \(n^2\): \[\frac{\ln n}{n^{2}+1} ≤ \frac{\ln n}{n^{2}}\] Now, we will check if the new series, \(∑_{n=2}^{\infty} \frac{\ln n}{n^{2}}\), converges by applying the Integral Test. Step 3: Apply the Integral Test

Step 3: Integral Test

To apply the Integral Test, we need to evaluate the following improper integral: \[\int_{2}^{\infty} \frac{\ln x}{x^2} dx\] First, we will use integration by parts to evaluate the integral. Let \(u = \ln x\) and \(dv = \frac{1}{x^2}dx\). Then, \(du = \frac{1}{x}dx\) and \(v = -\frac{1}{x}\). Now, we can apply integration by parts: \[\int_{2}^{\infty} \frac{\ln x}{x^2} dx = -\frac{\ln x}{x}\Big|_{2}^{\infty} - \int_{2}^{\infty} \frac{1}{x^2} dx\] The first term, \(-\frac{\ln x}{x}\Big|_{2}^{\infty}\), goes to zero as \(x\) goes to infinity. The second integral converges, as it is a p-series with p>1 (p=2): \[- \int_{2}^{\infty} \frac{1}{x^2} dx = -\left[-\frac{1}{x}\right]_{2}^{\infty} = 1/2\] Since the improper integral converges, we can conclude that \(∑_{n=2}^{\infty} \frac{\ln n}{n^{2}}\) converges by the Integral Test. Step 4: Determine if the original series is absolutely convergent, conditionally convergent, or divergent

Step 4: Determine convergence

Since the series \(∑_{n=2}^{\infty} \frac{\ln n}{n^{2}}\) converges, the absolute value of the original series must also converge. Therefore, the original series, \(∑_{n=2}^{\infty} \frac{(-1)^{n+1} \ln n}{n^{2}+1}\), is absolutely convergent.

Key Concepts

Absolute Convergence

Absolute convergence refers to when a series converges, even if we take the absolute value of each term. In simpler terms, consider a series with alternating positive and negative terms. By removing the alternating sign and checking if the resulting series converges, we determine its absolute convergence.
This means that if the series with absolute values also converges, the original series is said to be absolutely convergent.

• To find absolute convergence, calculate the absolute value of each term in the series.
• Try to determine if this new, non-alternating series converges.
• If it does, then the original series is absolutely convergent.
• Absolute convergence implies the series does not depend on the order of its terms for convergence.

In the given example, after taking the absolute value of the series, the sum \[∑_{n=2}^{fty} \frac{\ln n}{n^{2}+1}\]was determined to be convergent using the Comparison Test. Hence, confirming absolute convergence of the original series.

Comparison Test

The Comparison Test is a method used to determine series convergence by comparing it with another series that is known to converge or diverge. It is particularly helpful if we can find a series to compare against that mirrors the behavior of our series.
The idea is simple:

• Find a series with positive terms that you know converges or diverges.
• Compare your series term-by-term with this reference series.
• If your series has terms smaller than a converging series, it too converges.
• If it's larger than a diverging series, it diverges.

In the exercise, the given series' absolute form, \[\sum_{n=2}^{∞} \frac{\ln n}{n^{2}+1}\] was compared with \[\sum_{n=2}^{∞} \frac{\ln n}{n^{2}}\] . By observing that its terms are consistently smaller and that the latter converges, the original absolute series was shown to converge, thus aiding in determining absolute convergence.

Integral Test

The Integral Test is a technique that relates a series to an integral to ascertain its convergence. This test is effective for series with positive, continuous, and decreasing terms. By transforming a series into an integral, we can often use analytical integration methods to assess behavior.
Here's how it works:

• Take the series' equivalent function, replacing its index with a variable (often \(x\)).
• Evaluate the improper integral from your starting term to infinity.
• If the improper integral converges, so does the series.
• If it diverges, then the series diverges as well.

In the solution, an integral of the form \[\int_{2}^{∞} \frac{\ln x}{x^2} dx\] was used. Through integration by parts, this improper integral was computed to converge, indicating that the series \[\sum_{n=2}^{∞} \frac{\ln n}{n^{2}}\] is convergent. The result of this test was crucial for supporting the convergence claim made by the Comparison Test.

Alternating Series

Alternating series are those where the terms alternate in sign. Many series follow this pattern, especially when involving trigonometric or logarithmic expressions. The Alternating Series Test provides a confirmation for convergence under specific conditions, primarily dealing with behavior as the series progresses further with larger inductices.
To decide on the convergence of an alternating series:

• Check if the absolute value of terms decreases steadily.
• If terms reduce in magnitude over time and approach zero, the series may converge.
• Alternating Series Test focuses less on absolute values but rather the sequence of declining magnitudes.

For the given exercise,\[\sum_{n=2}^{∞} \frac{(-1)^{n+1} \ln n}{n^{2}+1}\] is alternating because of \((-1)^{n+1}\). However, to confirm absolute convergence, the steps involved directly checking for absolute and not conditional convergence, rendering the alternating nature secondary in this instance.