Problem 40 - Algebra and Trigonometry

Question

\(37-40\) . Find the first four partial sums and the nth partial sum of the sequence \(a_{m}\) \(a_{n}=\log \left(\frac{n}{n+1}\right) \quad[\text { Hint: Use a property of logarithms to }\) write the \(n\) th term as a difference.]

Step-by-Step Solution

Answer

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

Step 1: Identify the Sequence

Consider the given sequence \(a_n = \log \left( \frac{n}{n+1} \right)\). We need to express this sequence using properties of logarithms.

Step 2: Using Properties of Logarithms

Apply 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 3: Calculate the First Partial Sum \(S_1\)

The first partial sum is simply the first term: \(S_1 = a_1 = \log 1 - \log 2 = -\log 2\).

Step 4: Calculate the Second Partial Sum \(S_2\)

The second partial sum is \(S_2 = a_1 + a_2 = (\log 1 - \log 2) + (\log 2 - \log 3) = \log 1 - \log 3 = -\log 3\).

Step 5: Calculate the Third Partial Sum \(S_3\)

For the third partial sum, \(S_3 = a_1 + a_2 + a_3 = (\log 1 - \log 2) + (\log 2 - \log 3) + (\log 3 - \log 4) = \log 1 - \log 4 = -\log 4\).

Step 6: Calculate the Fourth Partial Sum \(S_4\)

For the fourth partial sum, \(S_4 = a_1 + a_2 + a_3 + a_4 = (\log 1 - \log 2) + (\log 2 - \log 3) + (\log 3 - \log 4) + (\log 4 - \log 5) = \log 1 - \log 5 = -\log 5\).

Step 7: General Formula for the nth Partial Sum \(S_n\)

Observe the pattern in the partial sums: each sum is progressively \(-\log k\) where \(k\) is increasing. Hence the \(n\)th partial sum is \(S_n = \log 1 - \log(n+1) = -\log(n+1)\).

Key Concepts

Properties of Logarithms

Logarithms are incredibly useful tools in mathematics, especially when dealing with multiplication and division. They convert multiplication into addition and division into subtraction, simplifying complex equations. One of the most important properties of logarithms is related to division, given by the formula:

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

This property allows us to break down a logarithm of a fraction into the difference of two logarithms. Thus, when faced with the sequence \( a_n = \log \left( \frac{n}{n+1} \right) \), we can rewrite it as \( a_n = \log n - \log(n+1) \). This manipulation simplifies further calculations, such as evaluating partial sums, due to the canceling effect it creates in a series.

Sequences and Series

A sequence is a list of numbers in a specific order determined by a pattern or rule. In this case, we have the sequence \( a_n = \log n - \log(n+1) \). A series, on the other hand, is the sum of the terms in a sequence. For any sequence, a partial sum is the sum of a specified initial part of the sequence.

• First Partial Sum: \( S_1 = a_1 = \log 1 - \log 2 = -\log 2 \)
• Second Partial Sum: \( S_2 = a_1 + a_2 = -\log 3 \)
• Third Partial Sum: \( S_3 = a_1 + a_2 + a_3 = -\log 4 \)
• Fourth Partial Sum: \( S_4 = a_1 + a_2 + a_3 + a_4 = -\log 5 \)

Each partial sum uses the properties of logarithms for simplification, revealing a clear pattern of negative logarithms that increases sequentially.

Mathematical Patterns

Mathematics often involves identifying and utilizing patterns. In this particular sequence and its partial sums, a distinct pattern emerges. Notice how each partial sum results in a term of the form \(-\log k\), where \(k\) increases consecutively, starting from 2.This pattern is valuable for predicting the behavior of the sequence without performing exhaustive calculations. By understanding this pattern, we find that the nth partial sum can be written as \( S_n = \log 1 - \log(n+1) = -\log(n+1) \). Recognizing and employing patterns is a fundamental skill that simplifies problem-solving in mathematics.