Problem 64 - Algebra and Trigonometry

Question

\(61-68\) 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 \( \sum_{n=2}^{100} (-1)^n \frac{1}{n \ln n} \).

Step 1: Identify the pattern

First, let's look at the given expression: \( \frac{1}{2 \ln 2} - \frac{1}{3 \ln 3} + \frac{1}{4 \ln 4} - \frac{1}{5 \ln 5} + \cdots + \frac{1}{100 \ln 100} \). Notice that it consists of alternating positive and negative terms, and the general term is of the form \( \frac{1}{n \ln n} \).

Step 2: Determine the general term

The terms alternate in signs based on even or odd n. For even \( n \), terms are positive, and for odd \( n \), terms are negative. We can express this alternation by using \( (-1)^n \). Thus, the general term is \[ (-1)^n \frac{1}{n \ln n}. \]

Step 3: Setup the sigma notation

The expression starts from \( n = 2 \) to \( n = 100 \). So, we use sigma notation to capture the entire expression as follows: \[ \sum_{n=2}^{100} (-1)^n \frac{1}{n \ln n}. \]

Step 4: Verify the setup

Ensure that the starting point \( n = 2 \) and the ending point \( n = 100 \) correctly encapsulate all terms of the series, respecting both the positive and negative alternations and the specific denominator in each term.

Key Concepts

Alternating Series

An alternating series is a type of infinite series where the signs of the terms switch back and forth between positive and negative. This type of series is often represented by having each term multiplied by \((-1)^n\) or \((-1)^{n+1}\). The use of \((-1)^n\) results in even terms being positive and odd terms being negative, while \((-1)^{n+1}\) would do the opposite.

The alternating nature of series means that it can converge to a limit, unlike some series that can grow indefinitely. Convergence of an alternating series is determined by the Alternating Series Test, which states that if the absolute value of the terms decreases steadily to zero, the series converges.

In the context of our problem, we identified the series \(\frac{1}{2 \ln 2} - \frac{1}{3 \ln 3} + \frac{1}{4 \ln 4} \cdots\) as an alternating series. Here, the \((-1)^n\) component makes sure that each term alternates its sign properly.

Logarithmic Functions

Logarithmic functions are mathematical functions related to the concept of exponents. The logarithm \(\ln n\) represents the power to which the base \(e\) (approximately 2.718) must be raised to produce the number \(n\). Logarithmic functions are very useful in solving equations involving exponential growth and decay.

In this exercise, the logarithm appears in the denominator of each term, specifically as \(\frac{1}{n \ln n}\). This means that each term is inversely related to both the natural number \(n\) and its natural logarithm, \(\ln n\).

The presence of \(\ln n\) generally causes the series to converge more slowly than it might otherwise. The ln function grows, but much slower than any polynomial function, which leads to smaller values of \(\frac{1}{n \ln n}\) as \(n\) increases.

General Term

The general term of a series is a mathematical expression that defines the nth term in the series in relation to n. General terms are critical as they standardize how any term in a series can be calculated and set the framework for using sigma notation.

Identifying the general term enables the reconstruction of the series. For this series, the term \(\frac{1}{n \ln n}\) served as the building block of the sequence. We included \((-1)^n\) to handle the alternating sign, producing the general term \((-1)^n \frac{1}{n \ln n}\).

Using the general term, we can express the series efficiently in sigma notation. This notation succinctly conveys the pattern and rule governing the series. Here, \(\sum_{n=2}^{100} (-1)^n \frac{1}{n \ln n}\) captures the entire series from the 2nd to the 100th term, considering both the alternation of signs and the logarithmic denominator.