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Magazine Chapter 11: Problem 17
Determine whether the series is convergent or divergent. $$ \sum_{n=1}^{\infty} \frac{1}{n^{2}+4} $$
Short Answer
Expert verified
The series is convergent.
Step by step solution
01
Recognize the Series Form
The given series is \( \sum_{n=1}^{\infty} \frac{1}{n^2+4} \). It resembles a general form of series \( \sum_{n=1}^{\infty} a_n \), where \( a_n = \frac{1}{n^2+4} \). To determine convergence, compare this with a known convergent or divergent series.
02
Comparison with the p-Series
Recall that the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \). Here, compare \( \frac{1}{n^2+4} \) with \( \frac{1}{n^2} \). Notice that for all \( n \geq 1 \), \( n^2 + 4 > n^2 \), so \( \frac{1}{n^2+4} < \frac{1}{n^2} \). Since \( \frac{1}{n^2} \) is a convergent p-series (with \( p = 2 \)), use the comparison test.
03
Apply the Limit Comparison Test
Firstly, find the limit: \[ \lim_{n \to \infty} \frac{\frac{1}{n^2+4}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n^2+4}. \] Simplify this to \( \lim_{n \to \infty} \frac{n^2}{n^2+4} = \lim_{n \to \infty} \frac{1}{1+\frac{4}{n^2}} = 1 \). Since the result is a finite, non-zero limit, the Limit Comparison Test tells us that both series either converge or diverge together.
04
Conclude Convergence
Since \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a convergent series, and our series \( \sum_{n=1}^{\infty} \frac{1}{n^2+4} \) was shown to behave similarly, the given series is also convergent according to the Limit Comparison Test.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding p-series
A p-series is a specific type of infinite series defined by the formula \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a constant. These series are interesting because their convergence depends entirely on the value of \( p \).
For p-series:
For p-series:
- If \( p > 1 \), the series converges.
- If \( p \leq 1 \), the series diverges.
The Comparison Test
The comparison test is a valuable tool when trying to determine whether a series converges or diverges. It involves comparing your series with another series whose convergence or divergence is already known. In this exercise, we compared the series \( \sum_{n=1}^{\infty} \frac{1}{n^2+4} \) with the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \).
Here is how the comparison test helped:
Here is how the comparison test helped:
- We noted that \( n^2 + 4 > n^2 \), which implies \( \frac{1}{n^2+4} < \frac{1}{n^2} \) for each term.
- Since \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a convergent p-series, by the comparison test, the original series must also converge.
Using the Limit Comparison Test
When direct comparison is difficult or not conclusive, the limit comparison test can be a powerful alternative. It involves taking the limit of the ratio of the terms from the two series.
Here's how it works:
Here's how it works:
- Find \( \lim_{n \to \infty} \frac{a_n}{b_n} \), where \( a_n \) and \( b_n \) are terms from the series we are investigating and the comparison series, respectively.
- If this limit is a positive, finite number, then both series either converge or diverge together.
Convergence and Divergence
The terms "convergence" and "divergence" describe the behavior of series. Determining whether a series converges or diverges is central in many areas of mathematics. Convergence means that as you add up an infinite number of terms, the sum approaches a finite number.
- If a series converges, it adds up to a specific, finite value as the number of terms goes to infinity.
- If a series diverges, the sum either increases indefinitely or fails to settle on any value as more terms are added.
