Problem 30 - Thomas Calculus

Question

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}(\sqrt{n}+1)} $$

Step-by-Step Solution

Answer

The series diverges by Comparison Test.

Step 1: Simplify the Series Term

The series given is \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}(\sqrt{n}+1)} \). We'll simplify each term \( \frac{1}{\sqrt{n}(\sqrt{n}+1)} \).Notice that \( \sqrt{n}(\sqrt{n}+1) = n + \sqrt{n} \). Therefore, each term of the series becomes \( \frac{1}{n + \sqrt{n}} \).

Step 2: Determine an Appropriate Test for Convergence

Knowing that the simplified term is \( \frac{1}{n + \sqrt{n}} \), we notice this looks similar to a \( \frac{1}{n} \) series. Therefore, the Comparison Test is quite suitable for this problem.

Step 3: Apply the Comparison Test

We compare \( \frac{1}{n + \sqrt{n}} \) to \( \frac{1}{n} \) because for large values of \( n \), \( n + \sqrt{n} \approx n \).Since for all \( n \geq 1 \): \[ n < n + \sqrt{n} \text{ implies } \frac{1}{n + \sqrt{n}} < \frac{1}{n} \]the given series is less than the harmonic series \( \sum \frac{1}{n} \), which diverges.

Step 4: Conclude with Divergence

Since \( \frac{1}{n + \sqrt{n}} < \frac{1}{n} \) and the harmonic series \( \sum \frac{1}{n} \) is known to diverge, the original series must also diverge by the Comparison Test.

Key Concepts

Comparison Test

The comparison test is a popular method for determining the convergence or divergence of infinite series. It involves comparing a given series with another series whose convergence behavior is already known. This is particularly useful when both series have terms that behave similarly as they approach infinity.

To use the comparison test, you usually compare your original series, say \( \sum a_n \), to another series \( \sum b_n \). There are a few good practices to keep in mind:

• If \( a_n \leq b_n \) for all \( n \), and \( \sum b_n \) converges, \( \sum a_n \) also converges.
• If \( a_n \geq b_n \) for all \( n \), and \( \sum b_n \) diverges, \( \sum a_n \) also diverges.

The idea is that if one series is smaller than a divergent series or larger than a convergent series, it must exhibit the same behavior as the comparison series.

In this particular problem, the series \( \sum \frac{1}{\sqrt{n}(\sqrt{n}+1)} \) was compared with the harmonic series \( \sum \frac{1}{n} \). We showed that each term of the original series is less than each term of the harmonic series, leading to the conclusion that the given series diverges.

Harmonic Series

The harmonic series is one of the most important series in calculus, and serves as a classical example of a divergent series. It is of the form \( \sum_{n=1}^{\infty} \frac{1}{n} \). Despite its simple appearance, the harmonic series diverges, meaning that its sum grows without bound as more terms are added.

It's useful to remember that, despite each term in the harmonic series (like \( \frac{1}{n} \)) decreasing in value, the series as a whole does not settle on a finite number. This is because the decrease in the magnitude of successive terms is too slow to offset their sum growing infinitely.

In many exercises, the harmonic series is used as a yardstick to measure whether other series diverge. Using the comparison test, as discussed, involves comparing a potentially unknown series to this well-studied divergent series to determine similar behavior in terms of convergence or divergence.

Divergence of Series

Determining whether a series diverges is a crucial part of understanding its behavior at infinity. Divergence implies that as more terms are added to the series, the sum grows indefinitely instead of settling on a number.

In general, for a series to diverge, it often contains terms that do not decrease to zero fast enough. Each term contributes a little too much, preventing the series from adding up to a finite value.

Specific tests, like the comparison test, help in confirming divergence. In our exercise, the series \( \sum \frac{1}{\sqrt{n}(\sqrt{n}+1)} \) was shown to diverge by comparing it to the harmonic series, which is known to diverge. It illustrates how similar behaviors between terms can be used as evidence for proving the divergence of other series. This conclusion showcases the power of divergence and comparison in calculus.