Problem 22 - Calculus: One and Several Variables

Question

Transform the first expression into the second by a change of indices. $$\sum_{n=2}^{12} \frac{(-1)^{n}}{n-1} ; \quad \sum_{i=1}^{11} \frac{(-1)^{k+1}}{k}$$

Step-by-Step Solution

Answer

The first expression \(\sum_{n=2}^{12} \frac{(-1)^{n}}{n-1}\) can be transformed into the second \(\sum_{i=1}^{11} \frac{(-1)^{i+1}}{i}\) by the change of indices \(i = n-1\).

Step 1: Identify the transformation

In this problem, the transformation is from n to i such that \(i = n-1\). This transformation shifts the index n by 1.

Step 2: Substitute the transformed index

Substitute the transformed index \(i = n - 1\) into the first expression: \(\sum_{n=2}^{12} \frac{(-1)^{n}}{n-1}\) becomes \(\sum_{i+1=2}^{i+1=12} \frac{(-1)^{i+1}}{(i+1)-1} = \sum_{i=1}^{11} \frac{(-1)^{i+1}}{i}\)

Step 3: Simplify the expression

The expression simplifies to \(\sum_{i=1}^{11} \frac{(-1)^{i+1}}{i}\), which is the second expression.

Key Concepts

Series Summation

Series summation is a powerful mathematical tool used to find the sum of terms within a sequence. Essentially, you take a series, which is a set of numbers following a specific pattern or rule, and sum them up to get a single numerical value. This concept is prevalent in various areas of mathematics, including calculus and number theory. A typical form of series summation appears as \[\sum_{n=a}^{b} f(n)\]where:

• \(n\) represents the index of summation, moving through values from the lower bound \(a\) to the upper bound \(b\).
• \(f(n)\) is the expression or function being summed.

The series is evaluated by substituting each integer value from \(a\) to \(b\) into \(f(n)\) and adding the results. For example, in the provided exercise, the original series, \(\sum_{n=2}^{12} \frac{(-1)^{n}}{n-1}\), sums the fraction terms derived using the formula from \(n = 2\) to \(n = 12\). Each term contributes to the total sum, and the series ends when the upper boundary is reached.

Index Shifting

Index shifting is a clever technique used to transform the indices of summation in a series without altering the overall value of the sum. This approach is particularly useful for changing the form of a series to make it easier to handle or compare with another expression. In the given problem, we shift the index \(n\) by subtracting 1 to form a new index \(i = n - 1\). This yields a new series:\[\sum_{i=1}^{11} \frac{(-1)^{i+1}}{i}\]What happens here is a straightforward transformation:

• Start with the original bounds: \(n = 2\) to \(n = 12\).
• Apply the transformation: \(i = n - 1\).
• New bounds become: \(i = 1\) to \(i = 11\).

The transformation simplifies the problem but maintains the integrity of the original series. It allows us to express the same sum with a potentially simpler or more convenient form, making computations or comparisons more straightforward.

Mathematical Transformation

Mathematical transformation involves changing the form or appearance of a mathematical expression while preserving its core properties or values. These transformations can simplify expressions, solve equations, or highlight certain features of the expression. In the context of our series, the transformation applied was an index shift. Here are some typical transformations:

• Changing variables (as seen with index shifting).
• Algebraically simplifying fractions or expressions.
• Applying trigonometric identities to simplify complex functions.

In this exercise, the index transformation allowed for a practical conversion of the original series expression. We effectively renamed \(n\) into \(i\) while adjusting the expression without losing or changing any values or relationships within the series. Such mathematical transformations are key to deepening our understanding and efficiency in solving complex problems, showcasing the inherent flexibility and power of mathematical notation and manipulation.