Problem 24 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{5 n^{3}-3 n}{n^{2}(n-2)\left(n^{2}+5\right)}\end{equation}

Step-by-Step Solution

Answer

The series converges by the p-series and Limit Comparison Test.

Step 1: Simplify the Series Expression

We begin by analyzing the given series \( \sum_{n=1}^{\infty} \frac{5n^3 - 3n}{n^2(n-2)(n^2+5)} \). Simplifying the expression involves rewriting the numerator and observing the dominant degree of \(n\). This can help us identify a potentially similar series. The numerator has a degree of 3 (the term \(5n^3\) dominates). Meanwhile, the denominator, when distributed, simplifies to a degree of 5 (the term \(n^5\) from \(n^2 \times n \times n^2\)). Thus, the dominant term of the series fraction becomes \( \frac{5n^3}{n^5} = \frac{5}{n^2} \).

Step 2: Compare with a Known Series

The series expression further simplifies to \( \frac{5}{n^2} \). Recognizing this as a \( p\)-series with \( p = 2 \), we know a \( p\)-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges when \( p > 1 \). Here, since \( p = 2 > 1 \), the series \( \sum_{n=1}^{\infty} \frac{5}{n^2} \) converges by the \( p\)-series test.

Step 3: Use the Limit Comparison Test

To confirm convergence, use the Limit Comparison Test with \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) (a convergent p-series). Compute the limit: \[ \lim_{n \to \infty} \frac{\frac{5n^3 - 3n}{n^2(n-2)(n^2+5)}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{5n^3 - 3n}{n^4(n-2)(n^2+5)} \times n^2 \]Simplifying:\[ \lim_{n \to \infty} \frac{5n^3 - 3n}{n^4 - 2n^3 + 5n^2} = \lim_{n \to \infty} \frac{5 - \frac{3}{n^2}}{n - \frac{2}{n} + \frac{5}{n^2}} \]For large \( n \), terms with \( \frac{1}{n} \) go to 0. Hence the limit is \( \frac{5}{n} \to 0 \) indicating the series converges.

Step 4: Conclude the Convergence or Divergence

Based on the comparison to the \( p \)-series and the Limit Comparison Test, we conclude that the series \( \sum_{n=1}^{\infty} \frac{5n^3 - 3n}{n^2(n-2)(n^2+5)} \) converges because its behavior at infinity reduces to that of the convergent \( p \)-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \).

Key Concepts

p-series

A p-series is an infinite series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. This type of series is a fundamental concept when studying convergence in calculus. The convergence or divergence of a p-series is determined solely by the value of \( p \).

Here’s how you can easily determine if such a series converges:

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

In our example from the exercise, the expression simplifies to \( \frac{5}{n^2} \), which is a p-series with \( p = 2 \). Since \( p = 2 > 1 \), this series converges according to the p-series test.

limit comparison test

The Limit Comparison Test is a useful tool in determining the convergence of a series, especially when you can compare it to a known convergent or divergent series. Here’s how it works:

• Take the series \( \sum_{n=1}^{\infty} a_n \) you want to examine and compare it to a known series \( \sum_{n=1}^{\infty} b_n \).
• Calculate the limit:
  \[ L = \lim_{n \to \infty} \frac{a_n}{b_n} \]

This test explains:

• If \( 0 < L < \infty \), then both series either converge or diverge together.
• If \( L = 0 \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( L = \infty \) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

In the given exercise, the limit comparison with \( \sum \frac{1}{n^2} \) shows \( L = 0 \), confirming the original series converges.

degree of polynomial

In the context of convergence tests, understanding the degree of a polynomial is crucial for simplifying and comparing series. The degree of a polynomial is the highest power of the variable in its expression. It tells us which term dominates the behavior as the variable becomes large.

For instance, in the exercise, the numerator \(5n^3 - 3n\) has a maximum degree of 3 because the largest power of \(n\) is 3. The denominator, \(n^2(n-2)(n^2+5)\), has a degree of 5 when expanded, with \(n \cdot n^2 \cdot n^2\) contributing the highest power. The fraction, therefore, behaves like \( \frac{1}{n^2} \) for large \(n\), which resembles a p-series with \(p=2\).

This simplification aided the recognition of a similar convergent series, leading to the use of the p-series test.

infinite series

An infinite series is a sum of infinitely many terms. It's usually expressed as \( \sum_{n=1}^{\infty} a_n \). We analyze whether this total converges to a finite limit or grows without bound. This is a central theme in mathematical analysis, often examined using various convergence tests.

There are key points to check whether an infinite series converges:

• See if it resembles a recognized convergent series, like a p-series or geometric series.
• Apply tests like the Limit Comparison Test, Ratio Test, or Integral Test.

Recognizing these common series patterns allows us to apply tests that lead to conclusions about convergence or divergence.

In practice, understanding these concepts enables mathematicians to deal with series sums in calculus and beyond, assuring precise calculations and solutions.