Problem 40 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The nth partial sum is \(-\log(n+1)\).

Step 1: 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.

Step 2: 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 \( \log(1) - \log(2) = 0 - \log(2) = -\log(2) \).

Step 3: 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, \( S_2 = \log(1) - \log(3) = -\log(3) \).

Step 4: 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 \( S_3 = \log(1) - \log(4) = -\log(4) \).

Step 5: 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 \( S_4 = \log(1) - \log(5) = -\log(5) \).

Step 6: 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 \( S_n = \log(1) - \log(n+1) = -\log(n+1) \).

Key Concepts

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.

This automatic reduction in complexity through subtraction makes telescoping series particularly interesting in mathematical analysis and problem-solving.

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: \( \log(a^b) = b\cdot\log(a) \)

In our sequence \( a_n = \log \left( \frac{n}{n+1} \right) \), the quotient property is used to separate the expression into the difference \( \log(n) - \log(n+1) \). This step is essential as it recognizes that each pair of logarithmic terms forms a difference and directly leads to the telescoping effect described earlier.Understanding these properties enables the simplification of complex series and sequences, making logarithmic operations more manageable.

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.

The partial sums \( S_1, S_2, S_3, \) and so on, are calculated by successively adding terms from the sequence. The elegance of this type of series lies in the telescopic nature, allowing a generalized form of partial sum \( S_n \) to be written simply as \( -\log(n+1) \).It is fascinating to see how such sequences transform and resolve through understanding their structure and applying series concepts.