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Chapter 7: Problem 60

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

Short Answer

Expert verified
The given series is divergent.

Step by step solution

01

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.
02

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\[ \lim_{{n \to \infty}} \frac{\frac{5}{n+\sqrt{n^{2}+4}}}{\frac{1}{n}} \]
03

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:\[ \lim_{{n \to \infty}} \frac{5}{1+\sqrt{1+0}} = \frac{5}{2} \]
04

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.

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(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0 . \overline{81} $$

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