Question

Use the Integral Test to determine the convergence of the given series. $$\sum_{n=1}^{\infty} \frac{n}{n^{2}+1}$$

Short answer

The series \( \sum_{n=1}^{\infty} \frac{n}{n^2 + 1} \) diverges by the Integral Test.

Step-by-step solution

Identify the Function for the Integral Test

The series given is \( \sum_{n=1}^{\infty} \frac{n}{n^2 + 1} \). For the Integral Test, we need a corresponding function \( f(x) \), where \( f(x) = \frac{x}{x^2 + 1} \) and the series term \( a_n = f(n) \). Ensure \( f(x) \) is positive, continuous, and decreasing for \( x \geq 1 \).

Check that the Function is Decreasing

To apply the Integral Test, \( f(x) = \frac{x}{x^2 + 1} \) must be decreasing. Compute the derivative \( f'(x) \) using the quotient rule. The derivative is \( f'(x) = \frac{(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2} \). For \( x \geq 1 \), \( 1 - x^2 \leq 0 \), hence \( f'(x) \leq 0 \) which confirms \( f(x) \) is decreasing.

Set Up the Integral

With \( f(x) = \frac{x}{x^2 + 1} \) decreasing, set up the integral \( \int_{1}^{\infty} \frac{x}{x^2+1} \, dx \) to evaluate convergence.

Calculate the Integral

Use substitution to evaluate \( \int \frac{x}{x^2+1} \, dx \). Let \( u = x^2 + 1 \), then \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2} du \). Substitute into the integral: \( \int \frac{x}{x^2+1} \, dx = \frac{1}{2} \int \frac{1}{u} \, du \). This results in \( \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|x^2+1| + C \).

Evaluate the Improper Integral

Evaluate the definite integral from 1 to infinity: \( \left. \frac{1}{2} \ln(x^2+1) \right|_{1}^{\infty} \). As \( x \to \infty \), \( \ln(x^2+1) \to \infty \), implying the integral diverges.

Conclude with the Integral Test

Since the integral \( \int_{1}^{\infty} \frac{x}{x^2+1} \, dx \) diverges, by the Integral Test, the series \( \sum_{n=1}^{\infty} \frac{n}{n^2 + 1} \) also diverges.

Key concepts

Convergence of Series

Understanding the concept of convergence in the context of series is essential in mathematical analysis. A series, which is the sum of an infinite sequence of terms, is said to converge if it approaches a specific value as more terms are added. Otherwise, it diverges. The Integral Test is one of many methods used to determine the convergence of a series. This method involves comparing the series with an improper integral.

The key requirement for the Integral Test is that the function associated with the series must be positive, continuous, and decreasing. If the improper integral of the function from a certain point to infinity converges to a finite value, then the series also converges. Conversely, if the integral diverges, so does the series.

In the given exercise, the series \( \sum_{n=1}^{\infty} \frac{n}{n^2+1} \) was tested for convergence using the Integral Test. By evaluating the associated function \( f(x) = \frac{x}{x^2+1} \) and finding that the corresponding integral diverges, it was concluded that the series itself diverges.

Improper Integrals

Improper integrals extend the concept of an integral to functions that are unbounded on the interval of integration or have unbounded intervals themselves. In such cases, traditional integral calculations are not directly applicable, and limits are used to define the integral.

An improper integral of the form \( \int_{a}^{\infty} f(x) \, dx \) is evaluated by considering \( \lim_{b \to \infty} \int_{a}^{b} f(x) \, dx \). If this limit exists and is finite, the improper integral converges. Otherwise, it diverges. In our exercise, the focus was on evaluating the integral \( \int_{1}^{\infty} \frac{x}{x^2+1} \, dx \).

Using substitution in the integration process allowed for calculating this integral, revealing its divergence. This divergence indicates that the area under the curve of \( f(x) = \frac{x}{x^2+1} \) from 1 to infinity is infinite, mirroring the behavior of the original series in the context of the Integral Test.

Decreasing Functions

The concept of decreasing functions is crucial when applying the Integral Test for convergence. A function \( f(x) \) is said to be decreasing on an interval if, as \( x \) increases within that interval, \( f(x) \) consistently decreases. Mathematically, for all \( a < b \), if \( f(a) \geq f(b) \), then \( f(x) \) is decreasing.

For the Integral Test, ensuring the function associated with the series is decreasing assures that the comparison between the series and the integral is valid. In our example, the function \( f(x) = \frac{x}{x^2+1} \) was examined for this property. Calculating the derivative \( f'(x) \) using the quotient rule showed that \( f'(x) \leq 0 \) for \( x \geq 1 \).

This mathematical confirmation of being decreasing is pivotal as it validates the preconditions required by the Integral Test, confirming that the test can be appropriately applied to determine the divergence of the series.