Chapter 9: Problem 9
Use comparison with \(\int_{0}^{\infty} 1 /\left(x^{2}+1\right) d x\) to show that \(\sum_{n=1}^{\infty} 1 /\left(n^{2}+1\right)\) converges to a number less than or equal to \(\pi / 2.\)
Short Answer
Expert verified
The series \( \sum_{n=1}^{\infty} \frac{1}{n^2+1} \) converges to a number \( \leq \frac{\pi}{2} \).
Step by step solution
01
Identify the Integral for Comparison
The given integral is \( \int_{0}^{\infty} \frac{1}{x^2 + 1} \, dx \). This is a convergent integral because it represents the arctangent function, which evaluates to \( \frac{\pi}{2} \) as \( x \to \infty \).
02
Set Up the Integral Comparison
We need to show that \( \sum_{n=1}^{\infty} \frac{1}{n^2 + 1} \) converges and its sum is less than or equal to the value of the integral \( \int_{0}^{\infty} \frac{1}{x^2 + 1} \, dx \). Notice that \( \frac{1}{n^2 + 1} \) is similar to the integrand \( \frac{1}{x^2 + 1} \).
03
Apply Integral Test for Convergence
The series \( \sum_{n=1}^{\infty} \frac{1}{n^2 + 1} \) can be compared term-by-term to the integral \( \int_{1}^{\infty} \frac{1}{x^2 + 1} \, dx \). By the integral test, because the integral of \( \frac{1}{x^2 + 1} \) from 1 to infinity converges, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2 + 1} \) also converges.
04
Bound the Series Under the Integral Result
Since the function \( \frac{1}{x^2+1} \) is decreasing, using the integral test \[ \sum_{n=1}^{\infty} \frac{1}{n^2+1} < \int_{1}^{\infty} \frac{1}{x^2+1} \, dx \] holds true. The integral evaluates to \( \frac{\pi}{4} \), and adding \( \frac{1}{0^2 + 1} = 1 \) below 1 gives a total contribution equal to \( 1 + \frac{\pi}{4} \leq \frac{\pi}{2} \). Thus, demonstrating the required inequality.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Test
The integral test is a powerful method used to determine whether a given infinite series converges or not. This test is particularly useful when dealing with series that behave in a similar way to a known, integrable function. For a series \( \sum_{n=1}^{\infty} a_n \), the integral test states that if \( a_n = f(n) \) for some positive, continuous, and decreasing function \( f(x) \), then the series converges if and only if the integral \( \int_{1}^{\infty} f(x) \, dx \) converges.
Here’s how it works:
Here’s how it works:
- Choose a function \( f(x) \) that matches the terms of the series, like \( f(n) = a_n \).
- Ensure that \( f(x) \) is positive, continuous, and decreases as \( x \) increases.
- Evaluate the improper integral \( \int_{1}^{\infty} f(x) \, dx \).
Comparison Test
The comparison test is a logical approach for determining the convergence or divergence of a series by comparing it to another series whose convergence behavior is already known. This comes in two forms: the direct comparison test and the limit comparison test.
Direct Comparison Test
Direct Comparison Test
- Compare the series \( \sum_{n=1}^{\infty} a_n \) to another series \( \sum_{n=1}^{\infty} b_n \), where each \( b_n \) is positive.
- If \( 0 \leq a_n \leq b_n \) for all \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
- If \( a_n \geq b_n \geq 0 \) and \( \sum b_n \) diverges, then \( \sum a_n \) diverges too.
- Take two positive sequences \( a_n \) and \( b_n \), and compute the limit \( L = \lim_{{n \to \infty}} \frac{a_n}{b_n} \).
- If \( L \) is a positive finite number, both \( \sum a_n \) and \( \sum b_n \) either converge or diverge together.
Improper Integrals
Improper integrals are a type of definite integral where either the interval of integration is infinite or the integrand approaches infinity within the interval. Handling such integrals requires a careful approach as they can represent limits themselves.
When evaluating an improper integral:
When evaluating an improper integral:
- If the limits of integration are infinite, like \( \int_{a}^{\infty} f(x) \, dx \), its value is found as a limit: \( \lim_{{b \to \infty}} \int_{a}^{b} f(x) \, dx \).
- If the function's value becomes unbounded, such as \( \int_{a}^{b} f(x) \, dx \) where \( f(x) \) has infinite discontinuities within \([a, b]\), then treat these points as limits.