Chapter 8: Problem 9
Determine the convergence of the given series using the Ratio Test. If the Ratio Test is inconclusive, state so and determine convergence with another test. $$\sum_{n=1}^{\infty} \frac{1}{n}$$
Short Answer
Expert verified
The series diverges. The Ratio Test is inconclusive, but the harmonic series is divergent.
Step by step solution
01
Identify the Series
The given series is the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \). Our task is to test its convergence using the Ratio Test.
02
Set Up the Ratio Test
The Ratio Test involves calculating the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), where \(a_n = \frac{1}{n}\). We need to substitute \( a_n \) and \( a_{n+1} \) into this limit.
03
Calculate \( a_{n+1} \) and \( \frac{a_{n+1}}{a_n} \)
\( a_{n+1} = \frac{1}{n+1} \) and compute \( \frac{a_{n+1}}{a_n} = \frac{1/(n+1)}{1/n} = \frac{n}{n+1} \).
04
Compute the Limit for the Ratio Test
Find \( L = \lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{n}{n+1} = 1 \).
05
Evaluate the Result of the Ratio Test
Since \(L = 1\), the Ratio Test is inconclusive. The test does not determine convergence if the limit equals 1.
06
Determine Convergence Using Another Test
The harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is known to be divergent. We know this from the Integral Test or previous knowledge. The series diverges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ratio Test
The Ratio Test is a popular method to determine the convergence or divergence of an infinite series. It's especially used when dealing with terms that contain factorials or exponential functions. To perform the Ratio Test, consider a series \(\sum a_n\):
- Calculate \( L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \)
- Evaluate \( L \):
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \) or \( L \) is infinite, the series diverges.
- If \( L = 1 \), the test is inconclusive.
harmonic series
The harmonic series is a classical example in the study of infinite series and divergence.It is defined as \( \sum_{n=1}^{\infty} \frac{1}{n} \).Despite its simple appearance, it has a fascinating property: it diverges.This means that as you add more and more terms, the sum grows without bound.
Here's why the harmonic series diverges:
Here's why the harmonic series diverges:
- The terms \( \frac{1}{n} \) do not approach 0 fast enough. In many convergent series, terms diminish quickly. For example, in a geometric series with \( r < 1 \), terms shrink exponentially.
- Grouping successive terms further illustrates this: For instance, \( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots \)
- Notice patterns like: \( \frac{1}{2} \), \( \frac{1}{3} + \frac{1}{4} > \frac{1}{2} \), \( \frac{1}{5} + ... + \frac{1}{8} > \frac{1}{2}, \ldots \) This process shows repeated additions of amounts greater than \( \frac{1}{2} \), hinting the series doesn’t sum to a finite number.
Integral Test
The Integral Test is another handy tool for determining the convergence of infinite series. This test is particularly useful for series where terms are function values of continuous, positive, decreasing functions.
To apply the Integral Test, take a series \( \sum_{n=1}^{\infty} a_n \) and relate it to the integral \( \int_{1}^{\infty} f(x) \, dx \) where \( f(n) = a_n \):
Evaluating \( \int_{1}^{\infty} \frac{1}{x} \, dx \):
To apply the Integral Test, take a series \( \sum_{n=1}^{\infty} a_n \) and relate it to the integral \( \int_{1}^{\infty} f(x) \, dx \) where \( f(n) = a_n \):
- If the integral converges, then the series converges.
- If the integral diverges, so does the series.
Evaluating \( \int_{1}^{\infty} \frac{1}{x} \, dx \):
- This integral yields \( \lim_{t \to \infty} \ln |t| - \ln |1| \).
- As \( t \to \infty \), the limit similarly diverges (it approaches infinity).