Problem 55 - Essential Calculus

Question

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{n+3}{n(n+2)} $$

Step-by-Step Solution

Answer

By using the Limit Comparison Test, we found that the given series diverges.

Step 1: Choose an appropriate comparison series

An appropriate comparison series b_n for this exercise would be 1/n as it has similar terms to our given series a_n = (n+3) / (n(n+2)).

Step 2: Calculate the limit of a_n / b_n

Determine the limit as n approaches infinity of (a_n / b_n). In this situation, a_n / b_n translates to \[\lim_{{n \to \infty}} \frac{(n+3) / (n(n+2))}{1/n}=\lim_{{n \to \infty}} \frac{n+3}{n^2+2n}\]. When calculating the limit of a rational expression as \( n \) approaches infinity, divide all terms by the highest power of \( n \) in the denominator, we get \[\lim_{{n \to \infty}} \frac{n/n+3/n^2}{n^2/n^2+2n/n^2}=\lim_{{n \to \infty}} \frac{1+3/n}{1+2/n}\]. As \( n \) approaches infinity, \( 3/n \) and \( 2/n \) approach zero. Therefore, the limit equals 1.

Step 3: Interpret the result according to the Limit Comparison Test

Since the limit of (a_n / b_n) as n approaches infinity equals 1, which is greater than zero, both series must either converge or diverge. Since our series for comparison 1/n is a known divergent series (Harmonic series), our given series also diverges.