Problem 70 - Precalculus: Mathematics for Calculus

Question

Write the sum using sigma notation. \(\frac{1}{2 \ln 2}-\frac{1}{3 \ln 3}+\frac{1}{4 \ln 4}-\frac{1}{5 \ln 5}+\cdots+\frac{1}{100 \ln 100}\)

Step-by-Step Solution

Answer

The sum is written as \( \sum_{n=1}^{99} (-1)^{n+1} \frac{1}{(n+1) \ln(n+1)} \).

Step 1: Identify the Pattern

Observe the terms given: \( \frac{1}{2 \ln 2}, -\frac{1}{3 \ln 3}, \frac{1}{4 \ln 4}, -\frac{1}{5 \ln 5}, \cdots, \frac{1}{100 \ln 100} \). The signs alternate and the denominator includes the natural logarithm of increasing integers starting from 2.

Step 2: Understand the Alternating Signs

Notice that the odd-term indices (e.g., 2, 4, etc.) have positive signs while the even-term indices (3, 5, etc.) have negative signs. This points to using \((-1)^{n+1}\) in the sum to control the sign alternation.

Step 3: Express Sum in General Term

From the pattern, the n-th term can be written as \((-1)^{n+1} \frac{1}{n+1 \ln(n+1)} \). Here, \(n\) starts at 1 and goes to 99 in order to cover terms from \(\frac{1}{2 \ln 2} \) to \(\frac{1}{100 \ln 100} \).

Step 4: Write in Sigma Notation

Using the general term, you can write the entire sum in sigma notation as: \[ \sum_{n=1}^{99} (-1)^{n+1} \frac{1}{(n+1) \ln(n+1)} \]

Step 5: Validate the Sequence

Double-check the formula by substituting a few values for \(n\). Plug in \(n = 1\) to get \(\frac{1}{2 \ln 2}\), for \(n = 2\) you get \(-\frac{1}{3 \ln 3}\), verifying the pattern and alternating signs are correctly implemented.

Key Concepts

Understanding Alternating Series

An alternating series is a sequence of numbers whose signs alternate between positive and negative. This can be seen in the given series: \( \frac{1}{2 \ln 2} - \frac{1}{3 \ln 3} + \frac{1}{4 \ln 4} - \frac{1}{5 \ln 5} + \cdots \). Alternating patterns are quite common in mathematical series. They often involve terms that change sign with each successive term. Identifying the pattern of signs is crucial to express the series succinctly using sigma notation. In this exercise, each term switches its sign from positive to negative and back again. This is achieved through a clever use of the expression \((-1)^{n+1}\), where \(n\) signifies the position of the term within the series. If \(n\) is even, the term becomes negative, and if \(n\) is odd, it turns positive. This manipulation of signs using exponents of -1 is a typical technique in managing alternating series behaviors. Understanding such operations is fundamental as they simplify the representation and comprehension of sequences that alternate, affecting the overall approach to solving exercises involving such scenarios.

Delving into the Natural Logarithm

The natural logarithm, often denoted as \(\ln(x)\), is the logarithm to the base \(e\). The number \(e\) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm is useful in modeling many natural phenomena and is an essential component in calculus and advanced mathematics.In the given series, each term involves a denominator that is a natural logarithm of a number. For example, terms like \( \ln 2\), \( \ln 3\), and \( \ln 4\) are seen. The role of the logarithm here is crucial as it determines how fast or slow the series converges. Generally, natural logarithms grow slower than linear functions, which influences the sequence's behavior.Knowing how natural logarithms work and their properties can help understand the intricacies of this series, especially regarding convergence and divergence when assessing larger sums or integrals over a defined range.

Analyzing the Sequence Pattern

Discerning the pattern in a sequence is often the first step in tackling series and sequences in mathematics. In this given problem, we have a sequence like: \( \frac{1}{2 \ln 2}, -\frac{1}{3 \ln 3}, \frac{1}{4 \ln 4}, -\frac{1}{5 \ln 5}, \cdots \). Analyzing such a sequence involves:

• Identifying the mathematical structure of each term, in this case, \( \frac{1}{n+1 \ln(n+1)} \).
• Understanding the range of numbers involved, starting from \(n=1\) to \(n=99\) for this series.
• Noticing the alternating nature by recognizing the impact of the \((-1)^{n+1}\) factor which controls each term's sign.

Breaking the sequence into valid steps allows for easier conversion into sigma notation, where you package this entire collection from \(n=1\) to \(n=99\) in a succinct mathematical representation. Such in-depth understanding aids not only in writing it in sigma notation but also in auditing and verifying each portion of the series for potential errors or logical missteps. Recognizing and documenting the pattern also helps in identifying if the sequence is part of a known mathematical series, which can then be manipulated or further simplified based on existing theorems.