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Chapter 8: Problem 31

Use the Comparison Test or Limit Comparison Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{1}{k^{3 / 2}+1}$$

Short Answer

Expert verified
If yes, why? Answer: Yes, the series converges. By applying the Limit Comparison Test with the comparison p-series $$\sum_{k=1}^{\infty} \frac{1}{k^{3/2}}$$, we find that the limit of the ratio of the two series as k approaches infinity is 1, which is a positive finite number. Since the comparison p-series converges (p = 3/2 > 1), the given series also converges.

Step by step solution

01

Determine the original series and choose a comparison series

The given series is: $$\sum_{k=1}^{\infty} \frac{1}{k^{3 / 2}+1}$$ Choose a comparison series. A good candidate to use is the p-series with p = 3/2: $$\sum_{k=1}^{\infty} \frac{1}{k^{3/2}}$$
02

Apply Limit Comparison Test

To determine the convergence of the given series using the Limit Comparison Test, we need to evaluate the limit of the ratio of the two series as k approaches infinity. Let $$a_k = \frac{1}{k^{3 / 2}+1}$$ and $$b_k = \frac{1}{k^{3/2}}$$, Compute the limit: $$ \lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\frac{1}{k^{3 / 2}+1}}{\frac{1}{k^{3/2}}}$$
03

Simplify the limit

To simplify the limit expression, we can multiply the numerator and denominator by \(k^{3/2}\): $$\lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{1}{k^{3 / 2}+1} \cdot \frac{k^{3/2}}{1} = \lim_{k \to \infty} \frac{k^{3/2}}{k^{3 / 2}+1}$$
04

Calculate the limit

To calculate this limit, divide both numerator and denominator by \(k^{3/2}\): $$\lim_{k \to \infty} \frac{k^{3/2}}{k^{3 / 2}+1} = \lim_{k \to \infty} \frac{1}{1 + \frac{1}{k^{3/2}}} = \frac{1}{1+0} = 1$$ Since the limit is a positive finite number (1 in this case), we can apply the Limit Comparison Test.
05

Conclude by using the Limit Comparison Test

Since the limit is 1, which is positive and finite, the given series converges if and only if the comparison series converges. The comparison series is a p-series with p = 3/2, which is greater than 1, and thus converges. Therefore, by the Limit Comparison Test, the given series 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
The convergence of series is a fundamental concept in calculus and analysis which involves determining whether the sum of an infinite number of terms reaches a finite value. There are various tests to establish convergence, one of which is the Limit Comparison Test. This test is particularly useful for series whose terms are not easy to compare directly to well-known convergent or divergent series.

In the context of this exercise, we looked at an infinite series and used the Limit Comparison Test to conclude that it converges. The test requires us to find a related series that we already know converges or diverges, and then compare the two series. If the limit of the ratio of their terms (as the series index goes to infinity) is finite and positive, both series will share the same convergence behavior.

It's crucial for students to understand this concept, as correctly identifying the behavior of series is not only essential for mathematics but also in applications such as physics and engineering where series can represent real-world phenomena.
p-Series
A p-series is an infinite series of the form \[\sum_{n=1}^\infty \frac{1}{n^p}\], where the variable 'n' denotes the term number, and 'p' is a constant. The convergence of a p-series depends on the value of 'p'. If 'p' is greater than 1, the series converges; if 'p' is less than or equal to 1, the series diverges.

In our exercise improvement advice, we utilized a p-series with \( p = \frac{3}{2} \) as a comparison series because it's known to converge. By comparing our original series to a known p-series, we could effectively determine the convergence of the original series without evaluating the entire sum, thus demonstrating the value of understanding p-series in the context of series convergence analysis.
Infinite Series
An infinite series is the sum of an infinitely large number of terms. It's represented as \[\sum_{n=1}^\infty a_n\], where \(a_n\) is the nth term in the series. Infinite series are encountered everywhere in mathematics, from basic calculus to more advanced topics.

The challenge in dealing with an infinite series is determining whether they converge to a specific limit or diverge to infinity (or fail to settle on a single value).

The series in our exercise, \(\sum_{k=1}^\infty \frac{1}{k^{3/2}+1}\), is an example of an infinite series. We can't simply add up an infinite number of terms, so instead, we use tests for convergence like the Limit Comparison Test to ascertain the behavior of these series as their number of terms grows without bound. Understanding the nature of infinite series and how they operate is crucial for grasping the overall concept of convergence and divergence within the study of series.

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Most popular questions from this chapter

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 \(10 ?\)

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

Consider the alternating series $$ \sum_{k=1}^{\infty}(-1)^{k+1} a_{k}, \text { where } a_{k}=\left\\{\begin{array}{cl} \frac{4}{k+1}, & \text { if } k \text { is odd } \\ \frac{2}{k}, & \text { if } k \text { is even } \end{array}\right. $$ a. Write out the first ten terms of the series, group them in pairs, and show that the even partial sums of the series form the (divergent) harmonic series. b. Show that \(\lim _{k \rightarrow \infty} a_{k}=0\) c. Explain why the series diverges even though the terms of the series approach zero.

The fractal called the snowflake island (or Koch island ) is constructed as follows: Let \(I_{0}\) be an equilateral triangle with sides of length \(1 .\) The figure \(I_{1}\) is obtained by replacing the middle third of each side of \(I_{0}\) with a new outward equilateral triangle with sides of length \(1 / 3\) (see figure). The process is repeated where \(I_{n+1}\) is obtained by replacing the middle third of each side of \(I_{n}\) with a new outward equilateral triangle with sides of length \(1 / 3^{n+1}\). The limiting figure as \(n \rightarrow \infty\) is called the snowflake island. a. Let \(L_{n}\) be the perimeter of \(I_{n} .\) Show that \(\lim _{n \rightarrow \infty} L_{n}=\infty\) b. Let \(A_{n}\) be the area of \(I_{n} .\) Find \(\lim _{n \rightarrow \infty} A_{n} .\) It exists!

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=\sqrt{2+a_{n}} ; a_{0}=1$$
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