Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} \frac{1}{n^{\ln 3}}$$

Short Answer

Expert verified
The series converges by the p-series test because \(\ln 3 > 1\).

Step by step solution

01

Understand the Series

The given series is \(\sum_{n=1}^{\text{∞}} \frac{1}{n^{\ln 3}}\). We need to determine whether this series converges or diverges.
02

Preliminary Test

The preliminary test involves checking whether the terms of the series approach zero as \(n \to \text{∞}\). Evaluate \(\lim_{{n \to \text{∞}}} \frac{1}{n^{\ln 3}}\). If this limit is not zero, the series diverges immediately.
03

Evaluate the Limit

Calculate the limit: \ \ \(\lim_{{n \to \text{∞}}} \frac{1}{n^{\ln 3}} = \lim_{{n \to \text{∞}}} n^{-\ln 3}\).\ Since \(\ln 3\) is a positive constant, \(n^{-\ln 3}\) will approach zero as \(n \to \text{∞}\). Thus, \( \lim_{{n \to \text{∞}}} \frac{1}{n^{\ln 3}} = 0\).
04

Select the Appropriate Test

Recognize that the series has the form \(\sum_{n=1}^{\text{∞}} \frac{1}{n^p}\). This suggests using the p-series test.
05

Apply the p-series Test

In a p-series \(\sum_{n=1}^{\text{∞}} \frac{1}{n^p}\), the series converges if \(p > 1 \) and diverges if \(p \leq 1\). Here, \(p = \ln 3\). Since \(\ln 3 \approx 1.0986\), which is greater than 1, the series converges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

series convergence tests
When working with series, it's essential to determine whether the series converges or diverges. This analysis helps us understand if the sum of the infinite series reaches a finite value or keeps growing indefinitely. Several tests can be applied to decide this.

The preliminary test is usually the first step. It checks whether the limit of the terms of the series approaches zero as the number of terms increases infinitely. If the terms do not approach zero, the series diverges automatically.

Beyond the preliminary test, there are several more specific tests. These include the p-series test, the ratio test, the root test, and the comparison test. Each has its criteria and can be useful depending on the series in question. Let's dive into the p-series test and the preliminary test in more detail.
p-series test
The p-series test is a straightforward and widely used test for determining the convergence or divergence of series in the form \( \sum_{{n=1}}^{{\text{∞}}} \frac{{1}}{{n^p}} \).

According to the p-series test, the series converges if the exponent \( p \) is greater than 1 and diverges if \( p \leq 1 \). For instance:
  • If \( p = 2 \), the series \( \sum_{{n=1}}^{{\text{∞}}} \frac{{1}}{{n^2}} \) converges.
  • If \( p = 0.5 \), the series \( \sum_{{n=1}}^{{\text{∞}}} \frac{{1}}{{n^{0.5}}} \) diverges.
In the given series example, we have \( \sum_{{n=1}}^{{\text{∞}}} \frac{{1}}{{n^{\text{{ln}} 3}}} \). Here, \( p = \text{{ln}} 3 \), which is approximately 1.0986. Since this value is greater than 1, the series converges by the p-series test. Understanding and applying the p-series test can make analyzing series simpler and more efficient.
preliminary test
The preliminary test is often the first method to apply when determining the convergence of a series. This test, also known as the nth-term test for divergence, involves analyzing the limit of the terms of the series as \( n \) tends to infinity.

The preliminary test states that if the limit of the terms \( a_n \) is not zero, then the series \( \sum a_n \) diverges. Mathematically, this is expressed as: \[ \text{{If }} \lim_{{n \to \text{∞}}} a_n \eq 0, \text{{ then }} \sum a_n \text{{ is divergent}}. \]

In the given problem, the terms of the series are \( \frac{{1}}{{n^{{\text{ln}} 3}}} \). We calculate the limit as \( n \) approaches infinity. Since \( \lim_{{n \to \text{∞}}} n^{-{\text{ln}} 3} = 0 \), we cannot conclude divergence from the preliminary test alone. This means we need additional tests, like the p-series test, to properly conclude the behavior of the series. The preliminary test, though simple, is a powerful starting point in the series analysis process.

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Most popular questions from this chapter

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges. $$\sum_{0}^{\infty} \frac{3^{n}(n+1)}{(x+1)^{n}}$$

Show that \(n ! > 2^{n}\) for all \(n > 3\). Hint: Write out a few terms; then consider what you multiply by to go from, say, \(5 !\) to \(6 !\) and from \(2^{5}\) to \(2^{6}\).

(a) Show that it is possible to stack a pile of identical books so that the top book is as far as you like to the right of the bottom book. Start at the top and each time place the pile already completed on top of another book so that the pile is just at the point of tipping. (In practice, of course, you can't let them overhang quite this much without having the stack topple. Try it with a deck of cards.) Find the distance from the right-hand end of each book to the right-hand end of the one beneath it. To find a general formula for this distance, consider the three forces acting on book \(n,\) and write the equation for the torque about its right-hand end. Show that the sum of these setbacks is a divergent series (proportional to the harmonic series). [See "Leaning Tower of The Physical Reviews," Am. J. Phys. 27, 121-122 (1959).] (b) By computer, find the sum of \(N\) terms of the harmonic series with \(N=25\) \(100,200,1000,10^{6}, 10^{100}\). (c) From the diagram in (a), you can see that with 5 books (count down from the top) the top book is completely to the right of the bottom book, that is, the overhang is slightly over one book. Use your series in (a) to verify this. Then using parts (a) and (b) and a computer as needed, find the number of books needed for an overhang of 2 books, 3 books, 10 books, 100 books.

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{3^{n}}{2^{2 n}}$$

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\frac{e^{x}}{1-x}$$

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