Question

Use the Direct Comparison Test to determine the convergence of the given series; state what series is used for comparison. $$\sum_{n=5}^{\infty} \frac{1}{\sqrt{n}-2}$$

Short answer

The series diverges using the Direct Comparison Test with \(\sum \frac{1}{\sqrt{n}}\).

Step-by-step solution

Reformulation of the Series

The given series is \( \sum_{n=5}^{\infty} \frac{1}{\sqrt{n} - 2} \). First, it is often useful to find a simpler series to compare it to. We observe that for large \(n\), \(\sqrt{n} - 2\) behaves similarly to \(\sqrt{n}\). Hence, \( \frac{1}{\sqrt{n} - 2} \approx \frac{1}{\sqrt{n}} \). We will use \( \sum_{n=5}^{\infty} \frac{1}{\sqrt{n}} \) for comparison.

Determine the Convergence of the Comparison Series

The comparison series is \( \sum_{n=5}^{\infty} \frac{1}{\sqrt{n}} \), which is a p-series with \( p = \frac{1}{2} \). Recall that a p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \). Here, since \( p = \frac{1}{2} \leq 1 \), we know that the series \( \sum_{n=5}^{\infty} \frac{1}{\sqrt{n}} \) diverges.

Apply the Direct Comparison Test

The Direct Comparison Test states if \(a_n \geq b_n\) for all \(n\) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges. Here, consider \( \frac{1}{\sqrt{n} - 2} > \frac{1}{\sqrt{n}} \) for \( n \geq 5 \), as subtracting 2 in the denominator reduces the value less than \( \sqrt{n} \), making the fraction larger. Since \( \sum_{n=5}^{\infty} \frac{1}{\sqrt{n}} \) diverges, so does \( \sum_{n=5}^{\infty} \frac{1}{\sqrt{n} - 2} \).

Key concepts

series convergence

Understanding series convergence is essential in calculus, especially when dealing with infinite series. In simple terms, a series converges when the sum of its terms approaches a finite number as more terms are added. Think of it like this: if you keep adding terms of the series, and their cumulative sum gets closer and closer to some fixed value, then the series converges. This is a critical concept since many mathematical problems can be expressed through summations, and knowing whether they converge helps in predicting behavior and outcomes. Conversely, when the sum grows indefinitely, we say the series diverges.

p-series

A p-series is a specific type of series represented as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. These series have unique convergence properties based on the value of \( p \). If \( p > 1 \), the series converges, meaning that it sums up to a finite number. However, if \( p \leq 1 \), the series diverges, meaning it does not sum up to a finite value and keeps growing. In the given exercise, the series \( \sum_{n=5}^{\infty} \frac{1}{\sqrt{n}} \) is a p-series with \( p = \frac{1}{2} \), which falls under the converges-if-\( p > 1 \) rule, hence it diverges.

divergence

Divergence occurs when an infinite series does not settle to a finite limit as the number of terms grows. Put differently, as you continue to add the terms of the series, the sum keeps increasing without ever stabilizing at a particular number. This is what happens in the exercise with the series \( \sum_{n=5}^{\infty} \frac{1}{\sqrt{n}} \), where the terms do not decrease fast enough for the series to converge. Recognizing divergence is vital as it indicates that the behavior we expect from a finite sum cannot be applied to an infinite setting.

Direct Comparison Test

The Direct Comparison Test is a powerful tool for determining the convergence or divergence of series by comparing them with other series whose convergence properties are already known. Essentially, you find a series that you know converges or diverges and compare its terms to the original series. If every term of your series \( a_n \) is larger than \( b_n \) from a divergent series \( \sum b_n \), then \( \sum a_n \) must also diverge. In the exercise, this test tells us that because \( \frac{1}{\sqrt{n} - 2} > \frac{1}{\sqrt{n}} \) for large enough \( n \), and the smaller series diverges, the original series must diverge as well.

comparison of series

Comparison of series involves evaluating an unfamiliar series by comparing it to another series with known convergence or divergence characteristics. This approach makes use of estimations and inequalities to handle series that are not straightforward to assess individually. In applying this method, look for a series that similarly mimics the behavior of your original series. Use inequalities to show whether elements of one are consistently larger or smaller than the other. By analyzing these relationships, as we did with \( \frac{1}{\sqrt{n} - 2} \) and \( \frac{1}{\sqrt{n}} \), you can conclude the convergence or divergence based on what you observe in the known series.