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Write the first four terms of the sequence \(\left\\{a_{n}\right\\}_{n=1}^{\infty}.\) $$a_{n}=\frac{(-1)^{n}}{2^{n}}$$

Short Answer

Expert verified
Answer: The first four terms of the sequence are \(\frac{-1}{2}, \frac{1}{4}, \frac{-1}{8},\) and \(\frac{1}{16}\).

Step by step solution

01

Calculate the first term (n=1)

To find the first term, substitute n=1 in the formula, \(a_{1}=\frac{(-1)^{1}}{2^{1}}\). This gives: $$a_{1}=\frac{-1}{2}$$
02

Calculate the second term (n=2)

To find the second term, substitute n=2 in the formula, \(a_{2}=\frac{(-1)^{2}}{2^{2}}\). This gives: $$a_{2}=\frac{1}{4}$$
03

Calculate the third term (n=3)

To find the third term, substitute n=3 in the formula, \(a_{3}=\frac{(-1)^{3}}{2^{3}}\). This gives: $$a_{3}=\frac{-1}{8}$$
04

Calculate the fourth term (n=4)

To find the fourth term, substitute n=4 in the formula, \(a_{4}=\frac{(-1)^{4}}{2^{4}}\). This gives: $$a_{4}=\frac{1}{16}$$ The first four terms of the sequence are \(\frac{-1}{2}, \frac{1}{4}, \frac{-1}{8},\) and \(\frac{1}{16}\).

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Key Concepts

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

Alternating Series
An alternating series is a sequence of numbers where the signs alternate between each term. This means one term is positive, and the next is negative, and so on. In mathematical terms, an alternating series can often be identified by the factor \((-1)^n\), which controls the sign of each term, alternating it with every increment of \(n\).
For example, if we look at the sequence given in the exercise, \(a_n = \frac{(-1)^n}{2^n}\), every odd term (when \(n = 1, 3, 5, \ldots\)) is negative, whereas every even term (when \(n = 2, 4, 6, \ldots\)) is positive.
This alternating behavior is a critical aspect that helps in the convergence analysis of infinite series, ensuring that the terms can average each other out to some extent.
  • Alternating series often help in approximating functions or integrals.
  • They can converge even if the individual terms don't tend toward zero very rapidly.
Fractional Expressions
Fractional expressions involve expressions that are ratios of two quantities, usually in the form \(\frac{a}{b}\). These are simple yet powerful components in mathematics and often form the basis of many algebraic and calculus operations.
In our sequence, \(a_n = \frac{(-1)^n}{2^n}\), each term is written as a fraction with a numerator and a denominator. This expression shows:
  • The numerator, \((-1)^n\), controls the sign of the term, making it positive or negative based on whether \(n\) is even or odd.
  • The denominator, \(2^n\), controls the size of the term, and as \(n\) increases, it causes the fraction to become smaller and smaller.
Understanding fractional expressions is crucial because they make it easy to see how each part of the term changes as \(n\) increases, especially in sequences and series. They can be used in breaking down complex mathematical concepts into simpler, more understandable parts.
Recurrence Relations
Recurrence relations are equations that recursively define sequences, where each term is a function of one or more of the previous terms. In other words, each term in the sequence can be expressed in terms of earlier ones, giving a stepping-stone approach to defining sequences. Although the sequence we are looking at is directly defined without a traditional recurrence relation, it lends itself to recursive thinking.
For example:
  • Recurrence relations are not directly visible, but can be inferred or constructed for analysis.
  • In some cases, defining a sequence through recurrence relations can make it easier to understand complex behaviors or patterns.
Usually, to establish a recurrence relation, you find the relation between terms like \(a_{n+1}\) and \(a_n\). However, in non-recursive definitions like \(a_n = \frac{(-1)^n}{2^n}\), no direct relation is given but can sometimes be inferred for pattern recognition and deeper insights.

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Most popular questions from this chapter

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5 7, \(11,13, \ldots .\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

Showing that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \operatorname{In} 1734,\) Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} .\) An elegant proof is outlined here that uses the inequality $$ \cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\left(\text { provided that } 0

Infinite products Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \ldots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \dots$$

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}}\). When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$ \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots $$ Use estimation techniques to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

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