Math Problem Statement

Determine if the series \( \sum_{n=1}^{\infty} \frac{7n^{3/2} + 4n}{9n^2 - 3n + 1} \) converges or diverges. State the test you will be using, use proper notation and show all of your work.

Solution

To determine whether the series

n=17n3/2+4n9n23n+1\sum_{n=1}^{\infty} \frac{7n^{3/2} + 4n}{9n^2 - 3n + 1}

converges or diverges, we will use the Limit Comparison Test by comparing this series with a simpler series. The Limit Comparison Test is particularly useful when the terms of the series have a similar form to a basic pp-series or a geometric series.

Step 1: Analyzing the Dominant Terms

The terms in both the numerator and denominator have leading powers that dominate the behavior of the fraction as nn \to \infty.

  • In the numerator, the dominant term is 7n3/27n^{3/2}.
  • In the denominator, the dominant term is 9n29n^2.

Thus, for large nn, the term

7n3/2+4n9n23n+17n3/29n2=791n1/2.\frac{7n^{3/2} + 4n}{9n^2 - 3n + 1} \approx \frac{7n^{3/2}}{9n^2} = \frac{7}{9} \cdot \frac{1}{n^{1/2}}.

So, we will compare the given series to the series

1n1/2.\sum \frac{1}{n^{1/2}}.

Step 2: Choosing the Comparison Series

The series 1n1/2\sum \frac{1}{n^{1/2}} is a pp-series with p=12p = \frac{1}{2}. Since p1p \leq 1, the pp-series 1n1/2\sum \frac{1}{n^{1/2}} diverges.

Step 3: Applying the Limit Comparison Test

To apply the Limit Comparison Test, we find the limit

limn7n3/2+4n9n23n+11n1/2=limn7n3/2+4n9n23n+1n1/2.\lim_{n \to \infty} \frac{\frac{7n^{3/2} + 4n}{9n^2 - 3n + 1}}{\frac{1}{n^{1/2}}} = \lim_{n \to \infty} \frac{7n^{3/2} + 4n}{9n^2 - 3n + 1} \cdot n^{1/2}.

Simplifying this expression:

=limn7n3/2n1/2+4nn1/29n23n+1=limn7n2+4n3/29n23n+1.= \lim_{n \to \infty} \frac{7n^{3/2} \cdot n^{1/2} + 4n \cdot n^{1/2}}{9n^2 - 3n + 1} = \lim_{n \to \infty} \frac{7n^2 + 4n^{3/2}}{9n^2 - 3n + 1}.

Now, divide the numerator and denominator by n2n^2:

=limn7+4n1/293n+1n2.= \lim_{n \to \infty} \frac{7 + \frac{4}{n^{1/2}}}{9 - \frac{3}{n} + \frac{1}{n^2}}.

As nn \to \infty, the terms 4n1/2\frac{4}{n^{1/2}}, 3n\frac{3}{n}, and 1n2\frac{1}{n^2} approach zero. Thus, we have

=79.= \frac{7}{9}.

Step 4: Conclusion

Since 79\frac{7}{9} is a positive, finite constant, the Limit Comparison Test tells us that the behavior of our series is the same as that of 1n1/2\sum \frac{1}{n^{1/2}}, which diverges. Therefore, the given series

n=17n3/2+4n9n23n+1\sum_{n=1}^{\infty} \frac{7n^{3/2} + 4n}{9n^2 - 3n + 1}

diverges.

Would you like further details or have any questions?

Here are some additional questions to deepen your understanding:

  1. What is the Limit Comparison Test, and how does it apply to series with polynomial terms?
  2. How do we determine the dominant term in a rational expression for series comparison?
  3. What is a pp-series, and how does its convergence depend on pp?
  4. Why is the Limit Comparison Test more effective than the Direct Comparison Test in some cases?
  5. How can you identify the convergence behavior of a series by just looking at the dominant terms?

Tip: When applying the Limit Comparison Test, focus on the highest power terms in both the numerator and the denominator to simplify your analysis.

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Math Problem Analysis

Mathematical Concepts

Series
Convergence and Divergence
Limit Comparison Test
p-Series

Formulas

Limit Comparison Test formula: \( \lim_{n \to \infty} \frac{a_n}{b_n} \)

Theorems

Limit Comparison Test
p-Series Test

Suitable Grade Level

College Calculus II or III