Chapter 9: Problem 35
(a) Find the partial sum, \(S_{n},\) of \(\sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right)\) (b) Does the series in part (a) converge or diverge?
Short Answer
Expert verified
(a) The partial sum is \( S_n = \ln(n+1) \). (b) The series diverges.
Step by step solution
01
Understand the Series
The given series is \( \sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right) \). This series can be rewritten using the properties of logarithms as \( \ln(n+1) - \ln(n) \). This revealing property indicates that it might be a telescoping series.
02
Identify the Telescoping Pattern
A telescoping series is one where successive terms cancel each other out when expanded. For our series, each term \( \ln(n+1) - \ln(n) \) results in cancellation where the \( \ln(n) \) of one term cancels with the \( -\ln(n) \) of the following term.
03
Formulate the Partial Sum, \( S_n \)
By summing \( \ln\left(\frac{n+1}{n}\right) \) from 1 to \( n \), the series telescopes:\[S_n = (\ln 2 - \ln 1) + (\ln 3 - \ln 2) + \ldots + (\ln(n+1) - \ln n) \]All intermediate terms cancel, leaving \( S_n = \ln(n+1) \).
04
Determine if the Series Converges or Diverges
Examine the progression of \( S_n = \ln(n+1) \). As \( n \to \infty \), \( S_n = \ln(n+1) \to \infty \). Thus, the partial sums tend towards infinity, indicating that the series diverges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Sum
A series' partial sum is the sum of the first few terms in that series. For example, if we have an infinite series made up of terms that follow a regular pattern, we can find the partial sum by adding up a finite number of these terms. In the case of our exercise, the series is given by \( \sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right) \).
- It initially seems complex due to the logarithmic form.
- However, by using the property of logarithms, \( \ln \left(\frac{n+1}{n}\right) \) can be expanded to \( \ln(n+1) - \ln(n) \).
Convergence and Divergence
In discussions of infinite series, two crucial concepts are convergence and divergence. These concepts help identify whether the series approaches a finite limit or not.For our series \( \sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right) \), the exploration starts with its partial sum \( S_n = \ln(n+1) \).
- To understand convergence, observe \( S_n \) as \( n \) becomes very large (approaches infinity).
- In this situation, \( \ln(n+1) \) grows higher (without bound).
Logarithmic Functions
Logarithmic functions play a central role in many mathematical explorations, including this exercise. A logarithm represents the power to which a number, known as the base, must be raised to produce another number. For example, in \( \ln(x) \), which is a natural logarithm, the base is \( e \) (approximately 2.718).In the given series \( \sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right) \), it was the properties of logarithms that allowed the expression to be simplified.
- Specifically, the quotients inside the logarithms unravelled due to the identity \( \ln \left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
- This simplification revealed a telescoping pattern.