Problem 60 - Essential Calculus

Question

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{5}{n+\sqrt{n^{2}+4}} $$

Step-by-Step Solution

Answer

The given series is divergent.

Step 1: Identification of a Comparable Series

We identify the simplest comparable series: \( \frac{1}{n} \) which is well characterized and we know that its behavior is a divergent series according to p-series test because p=1 which is less than or equals to 1 thus is divergent.

Step 2: Limit Comparison Test (LCT)

We proceed with the Limit Comparison Test. This involves computing the limit of the ratio of the terms of the two series as n approaches infinity. The Limit Comparison Test states that if the limit is a positive finite number, then both series have the same behavior (both converge or both diverge).Therefore,compute\[ \lim_{{n \to \infty}} \frac{\frac{5}{n+\sqrt{n^{2}+4}}}{\frac{1}{n}} \]

Step 3: Simplification of Limit

After cleaning up, the limit we have to compute becomes: \[ \lim_{{n \to \infty}} \frac{5}{1+\sqrt{1+\frac{4}{n^{2}}}} \]Since as n approaches infinity, \(\frac{4}{n^{2}}\) approaches 0, the limit simplifies to:\[ \lim_{{n \to \infty}} \frac{5}{1+\sqrt{1+0}} = \frac{5}{2} \]

Step 4: Conclusion based on LCT

The limit is a positive finite number (5/2). According to the Limit Comparison Test, this means the given series has the same behavior as the comparison series, \( \frac{1}{n} \), which is divergent. Therefore, the given series is also divergent.