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Chapter 11: Problem 26

Determine whether the series is convergent or divergent. $$ \sum_{n=1}^{\infty} \frac{n}{n^{4}+1} $$

Short Answer

Expert verified
The series is convergent.

Step by step solution

01

Analyze the General Term

The general term of the series is \( a_n = \frac{n}{n^4 + 1} \). As \( n \to \infty \), the denominator grows faster than the numerator because the highest degree term in the denominator is \( n^4 \), whereas in the numerator it's \( n^1 \).
02

Test for Divergence

According to the divergence test, if \( \lim_{n \to \infty} a_n eq 0 \), the series diverges. We have \( a_n = \frac{n}{n^4+1} \) which simplifies to \( \frac{1}{n^3} \) as \( n \to \infty \). Thus, \( \lim_{n \to \infty} a_n = 0 \), and the test is inconclusive.
03

Use the Limit Comparison Test

We will compare it with \( b_n = \frac{1}{n^3} \), a convergent p-series with \( p = 3 > 1 \). Compute \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n/(n^4+1)}{1/n^3} = \lim_{n \to \infty} \frac{n^4}{n^4+1} = 1 \).
04

Conclude with Limit Comparison Test Result

Since \( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 eq 0 \), by the limit comparison test, the behavior of \( \sum a_n \) is the same as \( \sum b_n \). Since \( \sum b_n = \sum \frac{1}{n^3} \) converges, so does \( \sum_{n=1}^{\infty} \frac{n}{n^4+1} \).

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a powerful tool when determining the convergence of series where at first glance, standard tests may not be directly applicable. This test is particularly useful because it allows you to compare a complicated series with a simpler, well-known series.
  • First, you're looking for a known comparison series, usually one with a p-series most easily analyzed.
  • To apply the test, compute the limit: \[L = \lim_{n \to \infty} \frac{a_n}{b_n}\]where \( a_n \) is your series term and \( b_n \) is from your comparison.
  • If \( 0 < L < \infty \), both series will either converge or diverge together.
Determining \( b_n \) involves simplifying your series' terms so you can identify familiar patterns, like polynomial terms that suggest a p-series. In our exercise, by comparing \( \frac{n}{n^4+1} \) with \( \frac{1}{n^3} \), the simplicity of \( \frac{1}{n^3} \) helps establish the convergence of the original series because \( L = 1 \).
P-series
A p-series is one of the most recognizable forms of an infinite series and it's characterized by its general term: \[\sum_{n=1}^{\infty} \frac{1}{n^p}\]where \( p \) is a positive constant.
  • If \( p > 1 \), the series converges.
  • If \( 0 < p \leq 1 \), the series diverges.
Understanding p-series is crucial because they serve as comparison standards in other tests, such as the Limit Comparison Test.In our example, we used the series \( \sum \frac{1}{n^3} \), a convergent p-series since \( p = 3 > 1 \). Knowing this helps us directly apply the Limit Comparison Test in making convergence judgments about more complex series.
Divergence Test
The Divergence Test, sometimes known as the nth-term test for divergence, is one of the simplest initial checks for series convergence, though it has clear limitations. This test states:
  • If \( \lim_{n \to \infty} a_n eq 0 \), the series \( \sum a_n \) diverges.
  • If \( \lim_{n \to \infty} a_n = 0 \), the test is inconclusive; the series still can diverge or converge.
It's a quick test to eliminate the possibility of convergence but doesn't guarantee it. In our scenario, the general term \( a_n = \frac{n}{n^4 + 1} \) simplifies to behave like \( \frac{1}{n^3} \) as \( n \to \infty \), with the limit equating to zero. This means reliance on further tests, like the Limit Comparison Test, was necessary to draw valid conclusions about convergence.

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Most popular questions from this chapter

Evaluate the indefinite integral as an infinite series. $$ \int x^{2} \sin \left(x^{2}\right) d x $$

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