Chapter 12: Problem 40
Find the first four partial sums and the \(n\) th partial sum of the sequence \(a_{n^{*}}\) \(a_{n}=\log \left(\frac{n}{n+1}\right) \quad[\text {Hint}:\) Use a property of logarithms to write the \(n\) th term as a difference.]
Short Answer
Expert verified
The nth partial sum is \(-\log(n+1)\).
Step by step solution
01
Understanding the Sequence
The sequence given is \( a_n = \log \left( \frac{n}{n+1} \right) \). Recognizing that it can be expressed as a difference of logarithms, we write this as \( a_n = \log(n) - \log(n+1) \). This will help simplify the calculation of partial sums by cancelling terms.
02
Calculate the First Partial Sum
The first partial sum, \( S_1 \), is simply the first term of the sequence: \( S_1 = \log \left( \frac{1}{2} \right) \). This is equivalent to \( \log(1) - \log(2) = 0 - \log(2) = -\log(2) \).
03
Calculate the Second Partial Sum
The second partial sum, \( S_2 \), is the sum of the first two terms: \( S_2 = a_1 + a_2 = \log \left( \frac{1}{2} \right) + \log \left( \frac{2}{3} \right) \). By applying the difference of logarithms, \( S_2 = \log(1) - \log(3) = -\log(3) \).
04
Calculate the Third Partial Sum
The third partial sum, \( S_3 \), is the sum of the first three terms: \( S_3 = a_1 + a_2 + a_3 = \log \left( \frac{1}{2} \right) + \log \left( \frac{2}{3} \right) + \log \left( \frac{3}{4} \right) \). This results in \( S_3 = \log(1) - \log(4) = -\log(4) \).
05
Calculate the Fourth Partial Sum
The fourth partial sum, \( S_4 \), is \( S_4 = a_1 + a_2 + a_3 + a_4 = \log \left( \frac{1}{2} \right) + \log \left( \frac{2}{3} \right) + \log \left( \frac{3}{4} \right) + \log \left( \frac{4}{5} \right) \). Simplifying, we find \( S_4 = \log(1) - \log(5) = -\log(5) \).
06
Generalize to the nth Partial Sum
The pattern we observe can be generalized for the \( n \)th partial sum: \( S_n = a_1 + a_2 + ... + a_n = \log \left( \frac{1}{2} \right) + \log \left( \frac{2}{3} \right) + ... + \log \left( \frac{n}{n+1} \right) \). By the telescoping nature of the series, this reduces to \( S_n = \log(1) - \log(n+1) = -\log(n+1) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Telescoping Series
A telescoping series is a unique type of sequence where many terms cancel each other out. This can happen when consecutive terms are arranged so that parts of the sequence get subtracted in a recurring pattern. In the provided exercise, the sequence is characterized as a telescoping series.Here's how it works: Each term in the sequence is expressed as a difference of logarithms, such as \( \log(n) - \log(n+1) \). What this means is that when we write out the series, consecutive terms like \( \log(n+1) \) from one term and \( -\log(n+1) \) from the next will cancel each other out.
- For the sequence \( a_n = \log \left( \frac{n}{n+1} \right) \), converting it into \( \log(n) - \log(n+1) \) demonstrates this property.
- This cancellation simplifies calculations significantly, as most terms vanish when summed over many elements.
Logarithmic Properties
Logarithmic properties play a crucial role in solving the given exercise by breaking down expressions for easier manipulation. Logarithms have several key properties which make them versatile tools in mathematics:
- Product Property: \( \log(a \times b) = \log(a) + \log(b) \)
- Quotient Property: \( \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \)
- Power Property: \( \log(a^b) = b\cdot\log(a) \)
Sequence and Series
A sequence is an ordered list of numbers following some specific rule, and a series is the sum of a sequence. When dealing with sequences, we often discuss partial sums—sums of the first few terms. These partial sums help understand the behavior of a sequence over time or as it approaches infinity.In the exercise, we are dealing with a specific sequence where each term involves the logarithm of a quotient. For sequences like this:
- Each term follows a predictable pattern, defined by \( a_n = \log \left( \frac{n}{n+1} \right) \).
- By summing these terms (forming a series), we can analyze how the sequence behaves as more terms are included.