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Chapter 9: Problem 22

Does the series converge or diverge? $$\sum_{n=0}^{\infty} \frac{2}{1+4 n^{2}}$$

Short Answer

Expert verified
The series converges by the Comparison Test.

Step by step solution

01

- Apply the Divergence Test

The first test to apply when analyzing the convergence of a series is the Divergence Test. Compute the limit of the general term as \( n \to \infty \): \[ \lim_{{n \to \infty}} \frac{2}{1+4n^2} = 0 \]If this limit is not zero, the series diverges. In this case, it is zero, so the Divergence Test is inconclusive, and we need to use another test.
02

- Consider a Comparison Test

Since the Divergence Test is inconclusive, consider using the Comparison Test. We need to find a series to compare with. Notice that \[ \frac{2}{1+4n^2} < \frac{2}{4n^2} = \frac{1}{2n^2} \] for all \( n \geq 1 \).
03

- Analyze the Comparison Series

The comparison series \( \sum_{n=1}^{ ext{鈭瀩} \frac{1}{2n^2} \) is a p-series with \( p=2 \), which converges (because \( p > 1 \)). Since \( \frac{2}{1+4n^2} \) is bounded above by a convergent series, we conclude the original series converges by the Comparison Test.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divergence Test
When you encounter a series, the first step in determining its convergence is often the Divergence Test.
This test helps you to make an initial assessment quickly. The idea here is straightforward: look at the limit of the series' general term as it approaches infinity.
If this limit is anything other than zero, then the series surely diverges. However, if the limit is zero, like in this example where: \[\lim_{{n \to \infty}} \frac{2}{1+4n^2} = 0\]it means the divergence test is inconclusive.
In such cases, the series may converge or diverge, and another test is required to come to a conclusion.
  • If the limit is zero, don't assume the series converges; it's not a sufficient condition.
  • If the limit is not zero, immediately conclude divergence.
Comparison Test
After the Divergence Test, one logical next step is the Comparison Test.
This test is all about comparing the series in question with another series whose convergence is well understood.
The key here is to find a suitable companion series. In our example, we compare\[\frac{2}{1+4n^2} \]with\[\frac{1}{2n^2}\]for all \( n \geq 1 \).
The comparison involves checking whether the original series is "smaller" or "larger."
Since we found that \(\frac{2}{1+4n^2} < \frac{1}{2n^2}\) it helps to ensure that the series is not growing too fast.
  • If the comparison series converges and the original series is smaller, the original series also converges.
  • If the comparison series diverges and the original series is larger, the original series diverges.
p-series
The p-series is a special type of series you'll often use for comparison purposes.
It's defined as:\[\sum_{n=1}^{\infty} \frac{1}{n^p}\]where \( p \) is a positive constant. What's pivotal here is the value of \( p \).
  • If \( p > 1 \), the p-series converges.
  • If \( p \leq 1 \), the p-series diverges.
In our scenario, the series \( \sum_{n=1}^{\infty} \frac{1}{2n^2} \) has \( p = 2 \), which is greater than 1.
Hence, it converges. Using this fact in combination with the Comparison Test, we deduced that our original series also converges. Understanding the behavior of p-series helps substantially when deciding which series to compare to.

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Most popular questions from this chapter

Are true or false. Give an explanation for your answer. \(\sum C_{n}(x-1)^{n}\) and \(\sum C_{n} x^{n}\) have the same radius of convergence.

Determine whether the series converges. $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} 2^{n}}{n^{2}}$$

Decide if the statements are true or false. Give an explanation for your answer. If all terms \(s_{n}\) of a sequence are less than a million, then the sequence is bounded.

Are true or false. Give an explanation for your answer. If a power series converges at one endpoint of its interval of convergence, then it converges at the other.

What values of \(a\) does the series converge? $$\sum_{n=1}^{\infty}\left(\frac{2}{a}\right)^{n}, a>0$$
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