Problem 62 - Algebra and Trigonometry

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

\( \sum_{n=2}^{100} (-1)^{n+1} \frac{1}{n \ln n} \)

Step 1: Identify the Pattern

First, observe the alternating sign pattern and the general term of the series. It can be seen that each term is of the form \( \frac{1}{n \ln n} \) where the sign alternates starting with a positive sign when \( n = 2 \).

Step 2: Establish the General Term

From the given series, the general form for each term is \( (-1)^{n+1} \frac{1}{n \ln n} \). Here, \( n \) starts at 2, since the first term is \( \frac{1}{2 \ln 2} \).

Step 3: Express with Sigma Notation

Now, express the sum using sigma notation. This will involve summing from \( n = 2 \) to \( n = 100 \). Thus, the series can be written in sigma notation as:\[\sum_{n=2}^{100} (-1)^{n+1} \frac{1}{n \ln n}\]

Step 4: Final Expression

The final expression using sigma notation is \( \sum_{n=2}^{100} (-1)^{n+1} \frac{1}{n \ln n} \), which encapsulates the entire series with alternating signs and terms of the form \( \frac{1}{n \ln n} \).

Key Concepts

Alternating Series

An alternating series is a special kind of series where the signs of the terms alternate between positive and negative. This means that each term's sign is the opposite of its predecessor. In the exercise given, you'll notice this pattern: the first term is positive, the second is negative, and this alternating pattern continues throughout.
The presence of alternating series often indicates a balance between terms. The positive terms add to the sum, while the negative terms reduce it. It's like a step forward, then one back, consistently throughout the series. This can sometimes lead to unique convergence properties where the sum of an infinite series can actually converge to a finite value, despite the infinite number of terms. Recognizing an alternating series quickly helps in identifying that pattern is crucial for manipulating and understanding series more deeply.

Pattern Recognition

Pattern recognition is like solving a puzzle to find the repeated structure within a sequence or series. In math, this skill is vital because it allows you to predict future elements based on the established ones.
In the given exercise, the pattern manifests in how each term follows a structure of the form \( \frac{1}{n \ln n} \). Moreover, the signs of the terms alternate, as recognized in the alternating series section. To efficiently use pattern recognition, break down the series to its most basic repeatable units.

• Look for repetitive terms or expressions.
• Identify if there's a particular mathematical operation applied consistently between terms (like '+1', or multiplication).
• Check for changing signs or harmonic patterns, especially in alternating series.

Pattern recognition gives you a tool to establish sequences easily, which is fundamental when writing series in compact forms such as the sigma notation.

General Term

The general term provides a blueprint for any term in a series or sequence. It outlines what mathematical operations are performed on \(n\) to yield any specific term. It's particularly powerful because once you determine the general term, it can be applied with any value of \(n\) to derive the corresponding term without needing to deduce each one individually.
From our exercise, the general term has been found to be \((-1)^{n+1} \frac{1}{n \ln n}\). This formula captures the alternating sign (using \((-1)^{n+1}\)) and the specific form of each term.

• Recognizing the general term simplifies complex series into understandable parts.
• It provides a clear method for expressing the series in sigma notation.
• Understanding the operations involved helps predict the behavior of terms beyond those explicitly stated.

Mastering the concept of a general term makes the manipulation of sequences and series into sigma notation seamless.

Logarithms

Logarithms are the inverse operations of exponentiation. In more straightforward terms, if you know that \(b^x = n\), a logarithm is used to express \(x\) as \(x = \log_b(n)\). It tells us "to what power must we raise \(b\) to get \(n\)?"
In the exercise, the logarithm appears in the denominator as \(\ln n\), which is the natural logarithm (logarithm to the base \(e\)). These are widely used in mathematical expressions due to their intrinsic properties of growth rate calculations and simplifying multiplicative processes to addition.
Logarithms in series have some important roles:

• They can transform expressions, simplifying the processes of multiplication to simpler additions.
• They are essential in solving equations where the unknown is an exponent.
• Understanding natural logarithms (\(\ln\)), often used in calculus, aids in understanding exponential growth or decay.

When dealing with series, especially ones involving calculus, being comfortable with logarithms and their properties is indispensable.