Problem 17
Question
Determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n^2 + 4} \)
Step-by-Step Solution
Verified Answer
The series \( \sum_{n=1}^{\infty} \frac{1}{n^2 + 4} \) is convergent.
1Step 1: Identify the series
The given series is \( \sum_{n = 1}^{\infty} \frac{1}{n^2 + 4} \). To analyze the convergence, we need to determine if this series converges or diverges as \( n \rightarrow \infty \).
2Step 2: Use the Comparison Test
To apply the comparison test, compare \( \frac{1}{n^2 + 4} \) with a known convergent series. Consider the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which is a p-series with \( p = 2 > 1 \), known to converge.
3Step 3: Establish the inequality
Observe that for \( n \geq 1 \), \( n^2 + 4 > n^2 \) implies \( \frac{1}{n^2 + 4} < \frac{1}{n^2} \). Since \( \frac{1}{n^2} \) is convergent and \( \frac{1}{n^2 + 4} \) is less than \( \frac{1}{n^2} \), we can apply the comparison test.
4Step 4: Apply the Comparison Test conclusion
Since \( \frac{1}{n^2 + 4} < \frac{1}{n^2} \) and \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is convergent, the Comparison Test concludes that \( \sum_{n=1}^{\infty} \frac{1}{n^2 + 4} \) is also convergent.
Key Concepts
Comparison TestP-SeriesInfinite Series
Comparison Test
The Comparison Test is a fascinating tool in determining the convergence or divergence of series. It allows us to draw parallels between a complex series and a simpler, well-known series. The core idea is to establish an inequality.
If a series we're interested in is smaller than a known converging series, it will converge too. Here's how it works:
If a series we're interested in is smaller than a known converging series, it will converge too. Here's how it works:
- Compare the terms of your given series with a target series of known behavior.
- If the terms of your series are less than the terms of a convergent series, your series will also converge.
- If the terms are larger than those of a divergent series, your series is divergent as well.
P-Series
A p-series is a well-studied type of series in mathematics, recognized by its simple structure. These series have the form: \[\sum_{n=1}^{\infty} \frac{1}{n^p} \]where \(p\) is a positive real number. Understanding p-series is crucial when using the Comparison Test because they often serve as the benchmark series:
- If \(p > 1\), the p-series converges. This is because the terms become significantly smaller, causing the sum to settle at a finite value.
- If \(p \leq 1\), the p-series diverges, meaning it grows without bound.
Infinite Series
An infinite series is essentially a sum that continues indefinitely, adding an infinite number of terms. The study of infinite series is critical in calculus and real analysis, as it helps in understanding how an infinite process can result in a finite sum.
For any infinite series, like:\[ \sum_{n=1}^{\infty} a_n \]convergence is determined by whether the sequence of partial sums approaches a finite limit as \( n \to \infty \). Key points regarding infinite series include:
For any infinite series, like:\[ \sum_{n=1}^{\infty} a_n \]convergence is determined by whether the sequence of partial sums approaches a finite limit as \( n \to \infty \). Key points regarding infinite series include:
- A series converges if its partial sums approach a particular number.
- A series diverges if the partial sums keep growing indefinitely or oscillate without settling.
- Tools like the Comparison Test enhance our ability to judge an infinite series' nature effectively.
Other exercises in this chapter
Problem 17
Test the series for convergence or divergence. $$ \sum_{n=1}^{\infty}(-1)^{n} \sin \left(\frac{\pi}{n}\right) $$
View solution Problem 17
Determine whether the series converges or diverges. \( \displaystyle \sum_{n = 1}^{\infty} \frac {1}{\sqrt {n^2 + 1}} \)
View solution Problem 17
Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum. \( 3 - 4 + \frac {16}{3} - {64}{9} + \cdot \cdot \cdot \)
View solution Problem 17
Find a formula for the general term \( a_n \) of the sequence, assuming that the pattern of the first few terms continues. \( \left\\{\begin{array} \frac {1}{2}
View solution