Problem 54 - Essential Calculus

Question

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=3}^{\infty} \frac{3}{\sqrt{n^{2}-4}} $$

Step-by-Step Solution

Answer

The given series diverges.

Step 1: Identify the comparable series

An obvious choice here as a comparison series would be \( \frac{1}{n} \) since it is simpler but still resembles the original series\( \frac{3}{\sqrt{n^{2}-4}} \). The series \( \frac{1}{n} \) is the harmonic series and it is known to diverge.

Step 2: Apply the limit comparison test

The Limit Comparison Test says that if \( \lim_{{n \to \infty}} \frac{a_{n}}{b_{n}} = c > 0 \), then the series \(\sum_{n=1}^{\infty}a_{n}\) and \(\sum_{n=1}^{\infty}b_{n}\) either both converge or both diverge. Here \(a_{n} = \frac{3}{\sqrt{n^{2}-4}}\) and \(b_{n} = \frac{1}{n}\). So compute the limit, \( \lim_{{n \to \infty}} \frac{a_{n}}{b_{n}} = \lim_{{n \to \infty}} \frac{n*3}{\sqrt{n^{2}-4}}. \)

Step 3: Simplify the limit

The fraction under the limit simplifies to \(3\frac{n}{\sqrt{n^{2}-4}}\), and taking limit as \(n \to \infty\), it approaches 3. Since the limit is a positive constant, according to the Limit Comparison Test, the given series has the same behavior as the series we compared it with.

Step 4: Apply the result of the Limit Comparison Test

The comparison series \( \frac{1}{n} \) is a diverging series (Harmonic series) and since we applied the Limit Comparison Test, it means that our original series \( \frac{3}{\sqrt{n^{2}-4}} \) also diverges.