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A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=0}^{\infty} x^{k}$$

Short Answer

Expert verified
Answer: The series converges for values of \(x\) in the interval \([0, 1)\).

Step by step solution

01

Identify the terms of the series

In this series, the terms are \(a_k = x^k\). Our task is to find the ratio \(\frac{a_{k+1}}{a_k}\) and apply the Ratio Test.
02

Apply the Ratio Test

According to the Ratio Test, we need to find the limit of the absolute value of the ratio between the terms \(a_{k+1}\) and \(a_k\). Let's calculate this ratio: $$\frac{a_{k+1}}{a_k} = \frac{x^{k+1}}{x^k} = x$$ Now we will take the limit of this expression as \(k\) approaches infinity: $$\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = \lim_{k \to \infty} |x| = |x|$$
03

Determine the convergence condition

According to the Ratio Test, the series converges if the limit calculated in Step 2 is less than 1. In this case, the convergence condition is \(|x| < 1\). Since we are only interested in non-negative values of x, we can write the condition as: $$0 \leq x < 1$$ So, the series converges for values of \(x\) in the interval \([0, 1)\).

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Key Concepts

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

Ratio Test
The Ratio Test is a powerful tool used in determining the convergence of infinite series. It is particularly useful for series where each term can be easily expressed or simplified using a quotient of successive terms. The concept revolves around assessing the behavior of the infinite series by looking at the limit of the ratio of successive terms.
To apply the Ratio Test, you first find the ratio between the \(k+1\)th term and the k-th term of the series, denoted as \(|\frac{a_{k+1}}{a_k}|\). If the limit of this ratio as \(k\) approaches infinity is less than 1, the series converges. Conversely, if this limit is greater than 1, the series diverges. If the limit equals 1, the test is inconclusive.
In our exercise, for the series \(\sum_{k=0}^{\infty} x^{k}\), the ratio simplifies to \(x\), and the convergence depends on the magnitude of \(x\) as determined by the Ratio Test.
Series Convergence
Series convergence is all about determining whether or not the sum of an infinite series can be expressed as a finite number. A series converges if its terms approach zero as you continue adding them indefinitely, leading to a finite sum. Otherwise, the series diverges, meaning it does not settle to any particular value.
To decide convergence, mathematicians use various tests, with the Ratio Test being a notable example. Convergence helps determine the behavior of series and functions expressed in terms of power series and ensures they are valid within their convergence intervals.
  • Convergent Series: Sum reaches a finite limit.
  • Divergent Series: Sum does not reach a finite limit.
Understanding convergence is essential in fields like calculus and analysis, especially when dealing with infinite processes and approximations.
Infinity Limit
The concept of the infinity limit often appears when analyzing series or sequences to determine their behavior as they extend indefinitely. It's a way of probing the trend as you approach boundlessness.
In the context of series, with the help of the ratio test, taking an infinity limit involves analyzing what happens as \(k\), the term number, tends to infinity. This step assures that the conditions hold as the sequence progresses indefinitely.
In our exercise, calculating \(\lim_{k \to \infty} |x|\) examines what happens to the ratio as \(k\) grows infinitely large. This step is crucial in confirming whether the series \(\sum_{k=0}^{\infty} x^{k}\) converges based on the value of \(x\). The continuity of \(x\) directly affects this limit, leading to an understanding of the convergence nature across different regions of \(x\).
Convergence Condition
The convergence condition is the requirement that must be satisfied for a series to converge. When employing the Ratio Test, this condition comes from comparing the limit of the ratio of consecutive terms to a threshold value of 1.
In the exercise for the series \(\sum_{k=0}^{\infty} x^{k}\), applying this condition leads us to conclude that the series converges for values where \(|x| < 1\). This essentially means the magnitude of \(x\) must be less than 1 for the sum of the series to remain finite and well-defined.
  • If \(\lim_{k \to \infty} |x| < 1\), the series converges.
  • If \(\lim_{k \to \infty} |x| > 1\), the series diverges.
  • The test is inconclusive when \(\lim_{k \to \infty} |x| = 1\).
For positive \(x\) values, this convergence condition translates to \(0 \leq x < 1\). Understanding these boundaries is paramount when working with power series, as they outline the region within which the series behaves predictably and results in a meaningful, finite answer.

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Most popular questions from this chapter

Suppose a ball is thrown upward to a height of \(h_{0}\) meters. Each time the ball bounces, it rebounds to a fraction r of its previous height. Let \(h_{n}\) be the height after the nth bounce and let \(S_{n}\) be the total distance the ball has traveled at the moment of the nth bounce. a. Find the first four terms of the sequence \(\left\\{S_{n}\right\\}\) b. Make a table of 20 terms of the sequence \(\left\\{S_{n}\right\\}\) and determine a plausible value for the limit of \(\left\\{S_{n}\right\\}.\) $$h_{0}=20, r=0.75$$

Evaluate the limit of the following sequences. $$a_{n}=\int_{1}^{n} x^{-2} d x$$

Explain the fallacy in the following argument. Let \(x=\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\) and \(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots .\) It follows that \(2 y=x+y\), which implies that \(x=y .\) On the other hand, $$x-y=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\cdots>0$$ is a sum of positive terms, so \(x>y .\) Thus, we have shown that \(x=y\) and \(x>y\).

Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than \(10^{-4} .\) Although you do not need it, the exact value of the series is given in each case. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2 k+1) !}$$

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\sqrt{2+a_{n}} ; a_{0}=1, n=0,1,2, \dots$$

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