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Chapter 1: Problem 11

Write the following series in the abbreviated \(\sum\) form. $$ \frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\cdots $$

Short Answer

Expert verified
\(\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2^{n+2}} \)

Step by step solution

01

Identify the pattern

Observe the given series to identify the pattern in the numerators and denominators. The denominators are powers of 2 and alternate between positive and negative signs.
02

Recognize the general term

The general term for the series can be written using a power of 2 in the denominator and alternation of signs (-1 raised to a power). The term in position n is \(\frac{(-1)^{n+1}}{2^{n+2}}\).
03

Write the series using summation notation

Using the general term identified in Step 2, write the series in summation notation: \[ \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2^{n+2}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinite Series
An infinite series is a sum of infinitely many terms. This means you keep adding terms forever. The series given in the exercise is an example of an infinite series: \(\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\text{...}\).
Infinite series can converge (sum up to a finite number) or diverge (sum up to infinity).
In this case, we deal with a convergent series because the terms get smaller and closer to zero as we proceed.
Alternating Series
An alternating series is a series where the terms change sign between positive and negative.
This unique property helps some series to converge even when individual terms are quite large.
In the example series: \( \frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\text{...} \), the signs alternate between positive and negative. This happens because of the \( (-1)^{n+1} \) term, making each next term have the opposite sign of the one before it.
General Term Identification
Identifying the general term of a series is crucial for writing its summation form.
You look for a pattern in the numerators and denominators of the sequence.
In the given series, the pattern is that each term's denominator is a power of 2 (2, 4, 8, 16, etc.).
For alternating signs, notice the terms alternate between positive and negative.
Combining these, the general term for our series is: \( \frac{(-1)^{n+1}}{2^{n+2}} \). Here, \( (-1)^{n+1} \) ensures alternation of signs and \( 2^{n+2} \) defines the sequence in the denominator.
Powers of 2
Powers of 2 are numbers in the form of \( 2^n \), like 1, 2, 4, 8, 16, and so on.
They're important in computer science and mathematics due to their use in binary systems and exponential growth modeling.
In our series, the denominators are all powers of 2: \( 4=2^2, 8=2^3, 16=2^4 \), etc.
Recognizing this pattern helps in defining the series's general term and writing it in summation notation.
Remember, understanding these powers allows you to simplify complex problems involving sequences and series.

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Most popular questions from this chapter

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\(\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n)^{3 / 2}}\)

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