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Chapter 7: Problem 55

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

Short Answer

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

Step by step solution

01

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)).
02

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.
03

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.

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