Problem 38 - Algebra and Trigonometry

Question

Find the first four partial sums and the \(n\)th partial sum of the sequence \(a_{n} .\) \(a_{n}=\log \left(\frac{n}{n+1}\right)\) [Hint: Use a property of logarithms to write the \(n\)th term as a difference.]

Step-by-Step Solution

Answer

The first four partial sums are \(-\log(2)\), \(-\log(3)\), \(-\log(4)\), \(-\log(5)\), and the nth partial sum is \(-\log(n+1)\).

Step 1: Write the term using logarithm properties

The sequence term is given by \( a_{n} = \log \left( \frac{n}{n+1} \right) \). We can use the property of logarithms that states \( \log \left( \frac{a}{b} \right) = \log(a) - \log(b) \). Therefore, \( a_{n} = \log(n) - \log(n+1) \).

Step 2: Find the first partial sum, \(S_1\)

Calculate the first partial sum by finding the sum of the first sequence term: \( S_1 = a_1 = \log(1) - \log(2) = -\log(2) \).

Step 3: Find the second partial sum, \(S_2\)

The second partial sum is calculated by adding the first two terms of the sequence: \( S_2 = a_1 + a_2 = (\log(1) - \log(2)) + (\log(2) - \log(3)) = -\log(3) \).

Step 4: Find the third partial sum, \(S_3\)

To find the third partial sum, add the first three sequence terms: \( S_3 = a_1 + a_2 + a_3 = (\log(1) - \log(2)) + (\log(2) - \log(3)) + (\log(3) - \log(4)) = -\log(4) \).

Step 5: Find the fourth partial sum, \(S_4\)

Add the first four sequence terms to get the fourth partial sum: \( S_4 = a_1 + a_2 + a_3 + a_4 = (\log(1) - \log(2)) + (\log(2) - \log(3)) + (\log(3) - \log(4)) + (\log(4) - \log(5)) = -\log(5) \).

Step 6: Determine the general formula for the nth partial sum, \(S_n\)

Observing the pattern, each subsequent term cancels out the log term from the previous partial sum, similar to a telescoping series. Hence, the \(n\)th partial sum is given by \( S_n = \log(1) - \log(n+1) = -\log(n+1) \).

Key Concepts

Logarithm Properties

Logarithms can often be simplified using their properties, which are crucial in solving mathematical problems involving such terms. One fundamental property is that the logarithm of a quotient can be expressed as the difference of two logarithms. Specifically, for any positive numbers \(a\) and \(b\), the property is written as:

• \(\log\left(\frac{a}{b}\right) = \log(a) - \log(b)\)

This property is particularly useful when dealing with sequences or series involving fractions, as it allows us to break down complex terms into simpler components. In the context of our problem, this property simplifies the term \(a_n = \log\left( \frac{n}{n+1} \right)\) to \(\log(n) - \log(n+1)\).
This simplification is key to understanding the behavior of the sequence and to computing the partial sums easily, as it sets up the cancellation effect observed in a telescoping series.

Telescoping Series

A telescoping series is a special type of series where much of the intermediate work cancels out, leaving a simple expression. This occurs thanks to the unique structure of the terms involved, allowing for a straightforward conclusion when finding sums.
In our original exercise, this telescoping nature becomes clear when calculating partial sums of the sequence \(a_n = \log(n) - \log(n+1)\). Each term effectively cancels out part of the adjacent terms:

• \(S_1 = \log(1) - \log(2)\)
• \(S_2 = (\log(1) - \log(2)) + (\log(2) - \log(3))\)
• \(S_3 = (\log(1) - \log(2)) + (\log(2) - \log(3)) + (\log(3) - \log(4))\)
• And so on...

Each step shows how subsequent \(\log\) terms cancel others out, leaving only the first and last terms of the sequence in the sum.
This pattern makes telescoping series a powerful tool for simplifying calculations and exploring sequences quickly.

Sequence

A sequence is an ordered list of numbers, each term generated by a specific formula. This formula determines the behavior and properties of the sequence. The sequence in this exercise is given by \(a_n = \log(n) - \log(n+1)\). Each term transforms based on its position, which is denoted by the index \(n\).
In sequences, understanding individual terms and their relationships is crucial when calculating partial sums. The term "partial sum" refers to the sum of the first few terms of a sequence, denoted commonly as \(S_n\). For our specific sequence, these partial sums reveal a pattern thanks to the telescoping property, resulting in the general formula \(S_n = -\log(n+1)\).
Working with sequences involves identifying patterns, leveraging properties like the logarithm properties, and recognizing structural patterns such as the telescoping nature, all leading to simpler mathematical solutions.