Problem 66 - Essential Calculus

Question

Test for convergence or divergence, using each test at least once. Identify which test was used. (a) \(n\) th-Term Test (b) Geometric Series Test (c) \(p\) -Series Test (d) Telescoping Series Test (e) Integral Test (f) Direct Comparison Test (g) Limit Comparison Test $$ \sum_{n=4}^{\infty} \frac{1}{3 n^{2}-2 n-15} $$

Step-by-Step Solution

Answer

As per the integral test, the series \(\sum_{n=4}^{\infty} \frac{1}{3 n^{2}-2 n-15}\) converges.

Step 1 - Identify the function

First, identify the function for the series. Here, \(f(n) = \frac{1}{3n^2 - 2n - 15}\) is the function.

Step 2 - Simplify the function

For convenience, it is often advisable to simplify the function if possible. Here, the function \(f(n) = \frac{1}{3n^2 - 2n - 15}\) can be simplified by factoring the denominator to be \(f(n) = \frac{1}{(3n-15)(n+1)}\).

Step 3 - Check the necessary conditions for Integral Test

Confirm that \(f(n)\) is positive, continuous, and decreasing for \(n ≥ 4\). Here, \(f(n)\) is positive as both \(n+1>0\) and \(3n-15>0\) for \(n ≥ 4\). As a rational function \(f(n)\) is continuous. Since the denominator increases as \(n\) increases, \(f(n)\) decreases. Thus, all necessary conditions for applying the Integral Test are confirmed.

Step 4 - Conduct the Integral Test

Evaluate the improper integral \(\int_{4}^{\infty} f(x) dx\). By substitution, we find that \(\int_{4}^{\infty} f(x) dx = \int_{4}^{\infty} \frac{1}{(3x-15)(x+1)} dx = \int_{4}^{\infty} \frac{1}{3x^2 - 2x - 15 } dx \). Calculating this integral yields a convergent result when the limits are applied.