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Use the test of your choice to determine whether the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$$

Short Answer

Expert verified
#Answer# The series $\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$ converges.

Step by step solution

01

Identify the series and corresponding function

The series is given as: $$\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$$ The corresponding function for the Integral Test is: $$f(k)=\frac{1}{k^{\ln k}}$$ Step 2: Setting up the integral
02

Set up integral

To apply the Integral Test, we want to evaluate the integral of the corresponding function from \(k=2\) to \(k=\infty\). Thus, we set up the integral as follows: $$\int_{2}^{\infty} \frac{1}{k^{\ln k}} dk$$ Step 3: Solving the integral
03

Solve the integral

To solve the integral, we will do a substitution. Let \(u = \ln k\). Then, \(du = \frac{1}{k} dk\). So when \(k=2\), \(u=\ln(2)\), and as \(k \to \infty\), \(u \to \infty\). Replacing \(k\) and \(dk\) in the integral, we get: $$\int_{\ln(2)}^{\infty} \frac{1}{e^{u^2}} du$$ Now, let's compare this integral with the integral of Gaussian function \(\int_{\ln(2)}^{\infty} e^{-u^2} du\). Since \(\frac{1}{e^{u^2}} = e^{-u^2}\), we have: $$\int_{\ln(2)}^{\infty} \frac{1}{e^{u^2}} du < \int_{\ln(2)}^{\infty} e^{-u^2} du$$ We know that the integral of Gaussian function converges; therefore, by Comparison Test, our integral also converges. Step 4: Conclusion
04

Conclude

Since the integral of the function converges, by the Integral Test, the original series converges as well. Thus, we can conclude that the series: $$\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$$ 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 a series is an essential topic in calculus. When we speak of convergence, we are looking to see whether the sum of infinite terms in a series approaches a finite number. If it does, we say the series converges. Otherwise, the series diverges. Testing for convergence often requires various methods and tests. One common approach is to apply the Integral Test. This involves translating the series into a continuous function and evaluating the improper integral over the desired limits. If the definite integral yields a finite result, then the series can be concluded to converge. Conversely, if the integral diverges, the series does as well. Understanding these principles helps in analyzing complex series and making sense of their behavior over infinity.
Comparison Test
The Comparison Test is a valuable tool for determining series convergence. It compares a given series to another series whose convergence properties are already known. There are two primary forms: - **Direct Comparison Test:** This involves directly comparing the terms of two series. If each term of your series is smaller than a converging comparison series, then your series also converges. - **Limit Comparison Test:** This form is useful when direct comparison isn't straightforward. By taking the limit of the ratio of the series terms as they approach infinity, we can determine convergence based on known series behavior. In our solution, the Comparison Test was used to compare the transformed integral to a Gaussian function's integral, which is known to converge. This verification finalizes that our integral and thus the original series converges as well.
Gaussian Function
The Gaussian function, characterized by its bell-like curve, plays a crucial role in mathematics and statistics. Its general form is given by the exponential function \( e^{-x^2} \). It is especially important in the context of series convergence because the integral of this function over the entire real line is known to converge. In our step-by-step solution, we observed this function's integral to draw conclusions about the original series' behavior. By known properties, the definite integral \( \int_{a}^{b} e^{-x^2} dx \) converges to a finite number, helping us conclude that similar integrals related to our series will also converge. This real-world utility of the Gaussian function aids significantly in confirming convergence characteristics in mathematical analysis.
Substitution Method in Integrals
The substitution method is a handy technique in calculus for simplifying integrals. Often, when dealing with complex expressions, we replace a part of the integrand with a single variable to transform the expression into a more manageable form. In the exercise at hand, substitution was employed by setting \( u = \ln k \), leading to \( du = \frac{1}{k} dk \). This change of variables turns the integral into a form reminiscent of the Gaussian integral, making the evaluation straightforward. Substitution not only simplifies integration but can also convert potentials of divergence into recognizable convergent forms, like in our exercise. Mastery of this method provides a foundation for solving a wide range of integrated problems effectively.

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Most popular questions from this chapter

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{n}{n^{2}+1}=0$$

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd .} \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N\), the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7\), and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. The Consumer Price Index (the CPI is a measure of the U.S. cost of living) is given a base value of 100 in the year \(1984 .\) Assume the CPI has increased by an average of \(3 \%\) per year since \(1984 .\) Let \(c_{n}\) be the CPI \(n\) years after \(1984,\) where \(c_{0}=100\)

Consider a wedding cake of infinite height, each layer of which is a right circular cylinder of height 1. The bottom layer of the cake has a radius of \(1,\) the second layer has a radius of \(1 / 2,\) the third layer has a radius of \(1 / 3,\) and the \(n\) th layer has a radius of \(1 / n\) (see figure). a. To determine how much frosting is needed to cover the cake, find the area of the lateral (vertical) sides of the wedding cake. What is the area of the horizontal surfaces of the cake? b. Determine the volume of the cake. (Hint: Use the result of Exercise 66.) c. Comment on your answers to parts (a) and (b).

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

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