Problem 20 - Calculus: Early Transcendentals

Question

Determine whether the series converges or diverges. \( \displaystyle \sum_{n = 1}^{\infty} \frac {n^2 + n + 1}{n^4 + n^2} \)

Step-by-Step Solution

Answer

The series converges.

Step 1: Analyze the General Term

Consider the general term of the series: \( a_n = \frac{n^2 + n + 1}{n^4 + n^2} \). To determine convergence or divergence, we can first simplify the expression for large \( n \).

Step 2: Simplify the General Term

For large values of \( n \), the dominant terms in the numerator and the denominator will be \( n^2 \) and \( n^4 \), respectively. Thus, we can approximate:\[a_n \approx \frac{n^2}{n^4} = \frac{1}{n^2}.\]This suggests the behavior of the series may resemble the \( p \)-series \( \sum \frac{1}{n^2} \) for large \( n \).

Step 3: Use the Limit Comparison Test

Compare \( a_n = \frac{n^2 + n + 1}{n^4 + n^2} \) with \( b_n = \frac{1}{n^2} \). Calculate the limit:\[\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n^2 + n + 1}{n^4 + n^2}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2(n^2 + n + 1)}{n^4 + n^2}.\]

Step 4: Simplify the Limit

Simplify:\[\frac{n^2(n^2 + n + 1)}{n^4 + n^2} = \frac{n^4 + n^3 + n^2}{n^4 + n^2}.\]Dividing the numerator and the denominator by \(n^4\):\[= \lim_{n \to \infty} \frac{1 + \frac{1}{n} + \frac{1}{n^2}}{1 + \frac{1}{n^2}}. \]As \( n \to \infty \), the fractions with \( n \) in the denominator approach zero, so:\[\lim_{n \to \infty} \frac{1 + \frac{1}{n} + \frac{1}{n^2}}{1 + \frac{1}{n^2}} = 1.\]

Step 5: Conclude using Limit Comparison Test

Since \( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 \) and \( \sum b_n = \sum \frac{1}{n^2} \) is a convergent \( p \)-series (where \( p = 2 > 1 \)), by the Limit Comparison Test, the series \( \sum a_n \) also converges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a handy tool when you want to determine the convergence or divergence of a series that looks similar to another series that you already know about. This test requires you to compare the terms of your series with the terms of a known series. If you can find a positive finite limit when you compare term ratios, you can conclude that both series behave the same way - they either both converge or both diverge.

Here's how it works:

• Identify the series you want to analyze, let's call it \(\sum a_n\).
• Choose a series \(\sum b_n\) that you know converges or diverges.
• Form the ratio \(\frac{a_n}{b_n}\) and take the limit as \(n\) approaches infinity.
• If \(\lim_{n \to \infty} \frac{a_n}{b_n} = c > 0\) and finite, then the series \(\sum a_n\) will converge or diverge if \(\sum b_n\) does likewise.

In our example, we used the Limit Comparison Test by taking \(b_n = \frac{1}{n^2}\), a classic convergent \(p\)-series, and found that the behavior of \(\sum a_n\) matched that of \(\sum b_n\).

p-series

A \(p\)-series is a special type of infinite series which has the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). Whether a \(p\)-series converges or diverges depends directly on the value of \(p\).

• If \(p > 1\), the series converges.
• If \(p \leq 1\), the series diverges.

Why is this important? When looking at complex series, you can sometimes simplify them to resemble a \(p\)-series. This makes analyzing their convergence much easier. In our original exercise, we compared the series to the \(p\)-series where \(p = 2\), which is known to be convergent.

These properties make \(p\)-series a crucial element in understanding series behavior and applying tests like the Limit Comparison Test effectively.

General Term Simplification

Simplifying the general term of a series is a critical step in identifying dominant terms which significantly affect a series' behavior for large values of \(n\). Complex series often contain terms of varying powers in both the numerator and the denominator. By focusing on dominant terms, you can simplify the expression, making it easier to compare with known series.

In our problem, \(a_n = \frac{n^2 + n + 1}{n^4 + n^2}\), the dominant term in the numerator is \(n^2\) and in the denominator is \(n^4\). For large \(n\), these dominate behavior, so by simplifying, we obtain \(a_n \approx \frac{1}{n^2}\). This visual simplification allows us to see that the series behaves like a known simple \(p\)-series, aiding in the application of the Limit Comparison Test.

Numerical Dominant Terms

Focusing on numerical dominant terms allows you to streamline complex expressions, which in turn helps identify the core behavior of a series. Typically, for large \(n\), only a few terms in a polynomial expression significantly impact the series' behavior. This is because powers of \(n\) rise sharply compared to lesser terms.

When determining convergence, look for:

• The largest power of \(n\) in the numerator and the denominator.
• Simplifying based on these terms for easier analysis.

In our example, the \(n^2\) and \(n^4\) are significant in their respective places, leading us to an approximated term of \(\frac{1}{n^2}\), aligning our series with a convergent \(p\)-series. Recognizing numerical dominance simplifies the problem, ensuring clarity when applying convergence tests.