Question

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

Short answer

The series converges.

Step-by-step solution

Understand the Integral Test

The Integral Test states that if \( f(n) = \frac{1}{n^2 + 1} \) is positive, continuous, and decreasing for \( n \geq 1 \), then the convergence of the series \( \sum_{n=1}^{\infty} f(n) \) is equivalent to the convergence of the improper integral \( \int_{1}^{\infty} f(x) \, dx \).

Verify the conditions for the Integral Test

First, check if \( f(n) = \frac{1}{n^2 + 1} \) is positive, continuous, and decreasing for \( n \geq 1 \). Since \( n^2+1 \) is always positive, \( f(n) \) is positive. It's continuous as the denominator never causes division by zero. To check if it's decreasing, note that as \( n \) increases, \( n^2+1 \) increases, making \( f(n) \) decrease.

Setup the improper integral

Now that the function meets the criteria, set up the improper integral that corresponds to the series: \( \int_{1}^{\infty} \frac{1}{x^2 + 1} \, dx \).

Compute the improper integral

The antiderivative of \( \frac{1}{x^2+1} \) is \( \arctan(x) \). Evaluate the integral:\[\int_{1}^{\infty} \frac{1}{x^2 + 1} \, dx = \lim_{b \to \infty} \left[ \arctan(x) \right]_{1}^{b}\]which simplifies to:\[\lim_{b \to \infty} \left( \arctan(b) - \arctan(1) \right)\]Since \( \arctan(b) \to \frac{\pi}{2} \) as \( b \to \infty \) and \( \arctan(1) = \frac{\pi}{4} \), the result is \( \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \).

Conclusion based on the Integral

Since the improper integral \( \int_{1}^{\infty} \frac{1}{x^2 + 1} \, dx \) converges to \( \frac{\pi}{4} \), the series \( \sum_{n=1}^{\infty} \frac{1}{n^2+1} \) also converges by the Integral Test.

Key concepts

Understanding Series Convergence

In mathematics, a series is a sum of terms of a sequence of numbers. Determining the convergence of a series means finding out whether the sum of its terms approaches a finite number as more and more terms are added. Series convergence is crucial in various fields, including calculus and statistics.
To assess convergence, there are several tests such as the Integral Test, Ratio Test, and comparison tests, each suitable for different types of series.

• The Integral Test, in particular, relates the series' convergence to the behavior of an associated improper integral.
• For the series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}+1}\), we use this test because it can provide a clear answer by evaluating the integral.

For convergence, all conditions of the test must be met: the function must be continuous, positive, and decreasing over its intended domain. If these conditions are satisfied, then the series converges if the related improper integral converges.

Exploring the Concept of Improper Integral

An improper integral is a type of definite integral where either the interval of integration is infinite, or the integrand approaches infinity within the interval.
This type of integrals often arises while evaluating series that extend to infinity or where functions have asymptotic behavior.

• In our example with the function \( f(x) = \frac{1}{x^2+1} \), we consider the integral \(\int_{1}^{\infty} \frac{1}{x^2+1} \, dx\).
• To check convergence, we compute this improper integral and determine whether it results in a finite value.

The process involves using limits to handle the infinity part:
Firstly, find the antiderivative of the function. Then use limits to calculate the integral's value as the upper bound approaches infinity.
By solving the integral and finding it converges to a finite number, we ascertain the related series also converges.

The Role of Antiderivatives

Antiderivatives serve as the cornerstone in solving integrals, especially when evaluating improper integrals for series tests.
Finding the antiderivative is akin to reversing differentiation—this process uncovers functions whose derivative gives the original function.

• Take, for example, \(\frac{1}{x^2+1}\), whose antiderivative is \(\arctan(x)\).
• Once the antiderivative is known, it assists in evaluating the definite integral, which in improper cases, involves limits.

The evaluation is typically expressed as a difference between the antiderivative values at specific bounds.
In our scenario:\[\lim_{b \to \infty} [\arctan(b) - \arctan(1)]\]helps finalize the integration. Changes thus captured in the antiderivative values reflect the "summed" behavior sustained over the interval.Ultimately, proving the integral converges ensures the series converges, completing the analysis efficiently.