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Choose your test Use the test of your choice to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{1}{k^{1+p}}, p>0$$

Short Answer

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Question: Determine the convergence of the following infinite series: $$\sum_{k=1}^{\infty} \frac{1}{k^{1+p}}, p>0$$ Answer: The given infinite series converges.

Step by step solution

01

Apply the Integral Test

To apply the Integral Test, we will consider the following improper integral: $$\int_1^\infty \frac{1}{x^{1+p}}dx$$ If this integral converges, so does the series, and if it diverges, the series also diverges.
02

Evaluate the integral

To solve the improper integral, evaluate the antiderivative and find the limit: $$\int_1^\infty \frac{1}{x^{1+p}}dx = \lim_{b \to \infty} \int_1^b \frac{1}{x^{1+p}}dx $$ We can evaluate the integral using power rule for integration: $$-\frac{1}{p}x^{-p}\Big|_{1}^{b} = \lim_{b \to \infty} \left[ -\frac{1}{p}b^{-p} + \frac{1}{p}\right]$$
03

Analyze the limit

To determine the convergence of the series, analyze the limit: \begin{equation} \lim_{b \to \infty} \left[ -\frac{1}{p}b^{-p} + \frac{1}{p}\right] \end{equation} Since \(p>0\), the term \(-\frac{1}{p}b^{-p}\) goes to 0 as b approaches infinity. Thus, the limit is: $$\lim_{b \to \infty} \left[ -\frac{1}{p}b^{-p} + \frac{1}{p}\right] = 0 + \frac{1}{p}$$ Since, the limit is a finite value, the improper integral converges.
04

Conclude about the series' convergence

As per the Integral Test, since the improper integral \(\int_1^\infty \frac{1}{x^{1+p}}dx\) converges, the given series: $$\sum_{k=1}^{\infty} \frac{1}{k^{1+p}}, p>0$$ also converges.

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Key Concepts

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

Convergence of Series
Understanding the convergence of series is crucial in mathematical analysis. A series is a sum of an infinite sequence of terms. Not all series converge, which means they do not always come to a finite sum. To determine whether a series converges, we can apply various tests, such as the Integral Test. If we find that the sum of the series approaches a specific number, then the series is said to converge. If it does not, the series diverges. In the exercise, we applied the Integral Test to determine the convergence of the series \( \sum_{k=1}^{\infty} \frac{1}{k^{1+p}} \), where \(p>0\). The outcome was that the series converges because the corresponding improper integral converged to a finite value.
Improper Integral
An improper integral is an integral with at least one of its limits extending towards infinity or having an integrand that approaches infinity within the interval of integration. In the discussed exercise, the improper integral was \( \int_1^\infty \frac{1}{x^{1+p}}dx \). To determine the convergence of this integral, we found its antiderivative and then evaluated the limit as the upper limit of integration approaches infinity. An improper integral converges when the limit of the integral as the upper boundary approaches infinity results in a finite value. In this case, because the limit was finite, we concluded that the improper integral and hence, the series, converges.
Power Rule for Integration
The power rule for integration is a key tool in calculus that simplifies the process of finding antiderivatives. It states that the integral of \(x^n\) with respect to \(x\) is \(\frac{x^{n+1}}{n+1}\), provided \(n eq -1\). In the context of our problem, we applied the power rule to find the antiderivative of \(\frac{1}{x^{1+p}}\). Although the function was presented in a fraction form, it could be rewritten as \(x^{-(1+p)}\). By applying the power rule, we derived the expression \(-\frac{1}{p}x^{-p}\), demonstrating how the power rule is a straightforward technique to solve integrals, and thus plays a crucial role in analyzing the convergence of our series.
Infinite Series
Infinite series are sums that contain an infinite number of terms. These series can either converge to a specific finite value or diverge, spreading out towards infinity or alternating indefinitely. The analysis of infinite series is a significant aspect of calculus and mathematical analysis because it helps in understanding the behavior of functions and sequences over an extensive range. In our given exercise, the series in question was \( \sum_{k=1}^{\infty} \frac{1}{k^{1+p}} \). It is an example of an infinite series because it has infinite terms starting from \(k=1\) up to infinity. Through application of tests like the Integral Test, mathematicians can determine whether such series converge or not, which has practical applications in various fields such as physics and engineering.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. A series that converges must converge absolutely. b. A series that converges absolutely must converge. c. A series that converges conditionally must converge. d. If \(\sum a_{k}\) diverges, then \(\Sigma\left|a_{k}\right|\) diverges. e. If \(\sum a_{k}^{2}\) converges, then \(\sum a_{k}\) converges. f. If \(a_{k}>0\) and \(\sum a_{k}\) converges, then \(\Sigma a_{k}^{2}\) converges. g. If \(\Sigma a_{k}\) converges conditionally, then \(\Sigma\left|a_{k}\right|\) diverges.

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. $$a_{n+1}=2 a_{n}\left(1-a_{n}\right) ; a_{0}=0.3, n=0,1,2, \dots$$

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. $$a_{n+1}=\frac{1}{2}\left(a_{n}+2 / a_{n}\right) ; a_{0}=2, n=0,1,2, \dots$$

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}$$

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