Problem 24 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=3}^{\infty} \frac{5 n^{3}-3 n}{n^{2}(n-2)\left(n^{2}+5\right)} $$

Step-by-Step Solution

Answer

The series converges by the Limit Comparison Test with a p-series.

Step 1: Analyze the General Term

The general term of the series is given by \( a_n = \frac{5n^3 - 3n}{n^2(n-2)(n^2+5)} \). Before proceeding with tests for convergence, it's often beneficial to simplify or approximate the term's behavior for large \( n \).

Step 2: Simplification of the General Term

Simplify \( a_n = \frac{5n^3 - 3n}{n^2(n-2)(n^2+5)} \). The leading term in the numerator is \( 5n^3 \) and in the denominator it is \( n^5 \) (from multiplying \( n^2(n^2) = n^4 \) and \( n-2 \) gives an additional factor of \( n \)). Thus, \( a_n \approx \frac{5n^3}{n^5} = \frac{5}{n^2} \).

Step 3: Apply the Comparison Test

Since \( a_n \approx \frac{5}{n^2} \) and this behaves like a \( p \)-series with \( p = 2 \), which is convergent (since \( p > 1 \)), we can use the Limit Comparison Test with \( b_n = \frac{1}{n^2} \).

Step 4: Limit Comparison Test Calculation

Calculate \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{5n^3 - 3n}{n^2(n-2)(n^2+5)}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{5n^3 - 3n}{n^2(n - 2)(n^2 + 5)} n^2 = \lim_{n \to \infty} \frac{5n^5 - 3n^3}{n^4(n - 2)(n^2 + 5)} \).

Step 5: Evaluate the Limit

Simplify \( \lim_{n \to \infty} \frac{5n^5 - 3n^3}{n^5} = \lim_{n \to \infty} \frac{5 - \frac{3}{n^2}}{1} = 5 \). Since this limit is a positive finite number, the series \( \sum_{n=3}^{\infty} \frac{5n^3 - 3n}{n^2(n-2)(n^2+5)} \) converges by the Limit Comparison Test.

Key Concepts

Comparison Test

The Comparison Test is a valuable tool used to determine the convergence or divergence of series, especially when a series doesn't easily display a p-series or geometric form. The fundamental idea is to compare the given series term \( a_n \) with another series, usually one that is well understood like a p-series or geometric series. If you can establish that \( a_n \) is always less than or equal to another series \( b_n \) and \( b_n \) converges, then \( a_n \) must also converge. Conversely, if \( a_n \) is always greater than or equal to \( b_n \) and \( b_n \) diverges, then \( a_n \) must also diverge.

• Choose a comparison series \( b_n \) that is easier to analyze.
• Use inequalities to establish relations between \( a_n \) and \( b_n \).
• Make sure the inequality holds for all relevant terms in the series.
• Conclude about the convergence or divergence based on these inequalities.

Remember: It's essential to choose a comparison series correctly and confirm the inequality for all terms in the series for the test to be valid.

Limit Comparison Test

The Limit Comparison Test extends the idea of the Comparison Test, allowing for a more straightforward analysis when direct inequality is challenging. The test involves calculating a limit to determine if two series \( a_n \) and \( b_n \) have the same convergence behavior. The series must have positive terms for this test to apply.To utilize the Limit Comparison Test, evaluate the limit \( \lim_{{n \to \infty}} \frac{{a_n}}{{b_n}} \). If this limit is a positive, finite number, both series either converge or diverge together. This process can simplify finding the convergence without needing an explicit inequality.

• Ensure each term \( a_n \) and \( b_n \) is positive.
• Choose a suitable comparison series \( b_n \).
• Calculate \( \lim_{{n \to \infty}} \frac{{a_n}}{{b_n}} \).
• Conclude based on whether the limit is a positive finite number.

This method is especially useful if the Comparison Test proves cumbersome or impractical.

P-Series

A p-series is a specific type of infinite series of the form \( \sum_{{n=1}}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The convergence of a p-series largely depends on the value of \( p \):

• The series converges if \( p > 1 \).
• It diverges if \( 0 < p \leq 1 \).

P-series are straightforward, but incredibly powerful in simplifying convergence problems. They often serve as the comparison series due to their clear convergence properties and are popular choices when applying the Comparison or Limit Comparison Tests.When handling series like \( \frac{5}{n^2} \), it resembles a basic p-series with \( p = 2 \). Here, since \( p > 1 \), we know this series converges, helping to compare and determine the convergence of more complex series by simplifying them into a p-series form.

Simplification of Terms

Simplification is crucial in solving series convergence problems as it offers clarity and allows effective use of convergence tests. Begin by identifying the dominant behavior of terms for large \( n \), which renders other minor terms insignificant in determining convergence.In many cases, like the series \( \frac{5n^3 - 3n}{n^2(n-2)(n^2+5)} \), simplifying reveals the dominant terms in both the numerator and denominator. By focusing on these dominant factors, the series approximates to \( \frac{5n^3}{n^5} = \frac{5}{n^2} \), fading into a simpler p-series form.

• Identify and simplify leading terms in numerator and denominator.
• Focus on asymptotic behavior as \( n \) approaches infinity.
• Use this simplified form in comparison with known series like a p-series.

This simplification tactic transforms cumbersome expressions into manageable forms, enabling straightforward application of comparison tests.