Problem 17 - Calculus: One and Several Variables

Question

Write the given sums as \(\sum_{k=3}^{10} a_{k}\) and as \(\sum_{i=0}^{7} a_{i+3}\) $$\frac{1}{2^{3}}+\frac{1}{2^{4}}-\cdots-\frac{1}{2^{10}}$$

Step-by-Step Solution

Answer

The given sum can be represented as \(\sum_{k=3}^{10} a_{k} = \sum_{k=3}^{10} (-1)^k \cdot \frac{1}{2^k}\) and \(\sum_{i=0}^{7} a_{i+3} = \sum_{i=0}^{7} (-1)^{i+3} \cdot \frac{1}{2^{i+3}}\).

Step 1: Rewrite the sum as \(\sum_{k=3}^{10} a_{k}\)

Since the terms follow a geometric progression with a common ratio of \(1/2\) and have alternating signs, we can represent the sum as: $$\sum_{k=3}^{10} a_{k} = \sum_{k=3}^{10} (-1)^k \cdot \frac{1}{2^k}$$

Step 2: Rewrite the sum as \(\sum_{i=0}^{7}a_{i+3}\)

Now, we want to rewrite the sum with a new index, \(i\), where \(k = i+3\). This means that when \(k=3\), \(i=0\) and when \(k=10\), \(i=7\). The new sum will become: $$\sum_{i=0}^{7} a_{i+3} = \sum_{i=0}^{7} (-1)^{i+3} \cdot \frac{1}{2^{i+3}}$$ Now we have represented the given sum in both requested forms: As \(\sum_{k=3}^{10} a_{k}\): $$\sum_{k=3}^{10} (-1)^k \cdot \frac{1}{2^k}$$ And as \(\sum_{i=0}^{7} a_{i+3}\): $$\sum_{i=0}^{7} (-1)^{i+3} \cdot \frac{1}{2^{i+3}}$$

Key Concepts

Alternating Series

An alternating series is a series where the signs of the terms alternately change. In other words, some terms are positive, while others are negative, following a regular pattern. This type of behavior is often denoted using \((-1)^k\), which means the sign of each term depends on whether the term's index, \(k\), is odd or even.
A classic example of an alternating series is:

• \( \sum_{k=3}^{10} (-1)^k \cdot \frac{1}{2^k} \)

When \(k\) is odd, \((-1)^k\) is negative, and the term is negative. When \(k\) is even, the term is positive.
This series is particularly interesting because it converges under certain conditions, like when the absolute value of the terms decreases monotonically and approaches zero. Alternating series are common in mathematical analysis and are used frequently to approximate functions or solve problems involving infinite sums.

Index Shifting

Index shifting is a technique used in mathematical series to change the index of summation. This allows us to represent the series in different forms without altering the value it sums to.

To perform index shifting, you define a new variable that relates to the original index. For example, in the original problem:

• With \(k = i + 3\), shifting the index means if \(k = 3\), \(i = 0\) and if \(k = 10\), \(i = 7\).

This transforms the sum \(\sum_{k=3}^{10} a_k\) into \(\sum_{i=0}^{7} a_{i+3}\). This method is powerful because it lets you manipulate sums into forms that are easier to work with or analyze, especially when dealing with more complex series or integrals.

Geometric Progression

A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio. Understanding this concept is crucial when dealing with series like the one given in the exercise.

In this example, the terms of the series are related by a geometric progression with a common ratio of \(\frac{1}{2}\). If the first term of our progression starts at \(\frac{1}{2^3}\), the next term is \(\frac{1}{2^4}\), and it continues this way.

• The sequence can be expressed as: \(\frac{1}{2^3}, \frac{1}{2^4}, \frac{1}{2^5}, \ldots\)
• The formula for the \(n\)-th term is \(\frac{1}{2^n}\)

This geometric nature allows for useful calculations such as finding the sum of the series, particularly helpful for problems involving convergence or alternating series where patterns emerge from the multiplication process.