Question

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

Short answer

The given series diverges.

Step-by-step solution

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.

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, \( \lim_{{n \to \infty}} \frac{a_{n}}{b_{n}} = \lim_{{n \to \infty}} \frac{n*3}{\sqrt{n^{2}-4}}. \)

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.

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.