Math Problem Statement

Use the Limit Comparison Test to determine the convergence or divergence of the series: ∑(n + 3) / (n^3 - 4n + 5) for n=1 to infinity.

Solution

To determine whether the given series converges or diverges using the Limit Comparison Test, we will follow these steps:

Series:

n=1n+3n34n+5\sum_{n=1}^{\infty} \frac{n + 3}{n^3 - 4n + 5}

Step 1: Identify a comparable series.

We begin by finding a simpler series to compare the given series to. Since the dominant term in the denominator is n3n^3 (as nn becomes large), we choose the simpler series: bn=1n2b_n = \frac{1}{n^2} This is because: n+3n34n+5nn3=1n2 as n\frac{n + 3}{n^3 - 4n + 5} \sim \frac{n}{n^3} = \frac{1}{n^2} \text{ as } n \to \infty

Step 2: Apply the Limit Comparison Test.

We compute the limit: L=limnn+3n34n+51n2L = \lim_{n \to \infty} \frac{\frac{n + 3}{n^3 - 4n + 5}}{\frac{1}{n^2}} Simplifying the expression: L=limn(n+3)n2n34n+5L = \lim_{n \to \infty} \frac{(n + 3) n^2}{n^3 - 4n + 5} As nn \to \infty, the highest powers of nn dominate, so we approximate the limit: L=limnn3+3n2n34n+5=limnn3(1+3n)n3(14n2+5n3)L = \lim_{n \to \infty} \frac{n^3 + 3n^2}{n^3 - 4n + 5} = \lim_{n \to \infty} \frac{n^3(1 + \frac{3}{n})}{n^3(1 - \frac{4}{n^2} + \frac{5}{n^3})} For large nn, the terms 3n\frac{3}{n}, 4n2\frac{4}{n^2}, and 5n3\frac{5}{n^3} tend to zero, so: L=11=1L = \frac{1}{1} = 1

Step 3: Conclusion based on the Limit Comparison Test.

Since L=1L = 1 and 0<L<0 < L < \infty, the Limit Comparison Test tells us that the behavior of the given series is the same as that of the simpler series 1n2\sum \frac{1}{n^2}. The series 1n2\sum \frac{1}{n^2} is a p-series with p=2>1p = 2 > 1, which converges.

Thus, the given series: n=1n+3n34n+5\sum_{n=1}^{\infty} \frac{n + 3}{n^3 - 4n + 5} converges by the Limit Comparison Test.


Would you like more details or have any questions?

Here are five related questions to explore further:

  1. How does the Limit Comparison Test differ from the Direct Comparison Test?
  2. What are the conditions for using the Limit Comparison Test?
  3. Can the Limit Comparison Test be applied if L=0L = 0?
  4. How would the series behave if the denominator's degree was less than 3?
  5. What is the convergence behavior of p-series when p1p \leq 1?

Tip: When using the Limit Comparison Test, always ensure that both series you're comparing are positive and behave similarly as nn approaches infinity.

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Math Problem Analysis

Mathematical Concepts

Limit Comparison Test
Series Convergence
p-Series

Formulas

L = lim (n -> ∞) (a_n / b_n)
a_n = (n + 3) / (n^3 - 4n + 5)
b_n = 1 / n^2

Theorems

Limit Comparison Test
p-Series Test

Suitable Grade Level

College/Advanced High School