/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Evaluate each series or state th... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each series or state that it diverges. $$\sum_{k=0}^{\infty} \frac{1}{16 k^{2}+8 k-3}$$

Short Answer

Expert verified
Answer: Converge

Step by step solution

01

Simplify the general term

We have the general term: $$\frac{1}{16 k^{2}+8 k-3}$$. Factor out \(16\) from the quadratic term: $$\frac{1}{16( k^{2}+\frac{1}{2}k-\frac{3}{16})}$$. We need to complete the square in order to further simplify the expression. The square root of the first term, k, must also be half the square root of the middle term: $$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$. Add and subtract the square of the half of the middle term's coefficient, we get: $$\frac{1}{16}\left[ \frac{1}{(k+\frac{1}{4})^2 - \frac{25}{16}} \right]$$
02

Apply the Comparison Test

Now that the general term has been simplified, we can apply the Comparison Test. To do so, we must find a series that converges and whose summand is greater than or equal to the given general term for all k in the domain. Given that the summand is a rational function, we can compare it to the summand of the p-series: $$\frac{1}{k^2}$$. The p-series with $$p = 2$$ is known to converge. Now, show that $$\frac{1}{16(k^{2}+\frac{1}{2}k-\frac{3}{16})} \leq \frac{1}{k^2}$$ for all $$k \geq 1$$. Note that for $$k \geq 1$$: $$16 k^{2} + 8 k - 3 \geq k^2$$ since $$(16k^2 + 8k - 3) \geq k^2$$. As a result, the given series satisfies the conditions of the Comparison Test and can be bounded by the convergent series $$\sum_{k=1}^{\infty} \frac{1}{k^{2}}$$. Therefore, we could conclude that the given series converges.

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Key Concepts

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

Comparison Test
The Comparison Test is a method used to determine the convergence or divergence of an infinite series. It works by comparing the series in question to another series whose convergence properties are already known. This test is particularly useful when dealing with complex series because it allows us to simplify our work by utilizing known series.To apply the Comparison Test, you find a series \(_k\) such that for all positive integers \(k\), each term \(a_k\) of the series \(_k\) can be compared to each term \(a_k\) of the series we are initially examining. Specifically:
  • If \(0 \leq a_k \leq b_k\) and \(_k\) converges, then \(\sum a_k\) also converges.
  • If \(a_k \geq b_k \) and \(_k\) diverges, then \(\sum a_k\) also diverges.
This step is essential, as demonstrated in the provided exercise solution where the target series was compared with a known \(p\)-series.
p-series
In mathematics, a \(p\)-series is an important concept when discussing series convergence. A \(p\)-series is a series of the form \[\sum_{k=1}^{\infty} \frac{1}{k^p}\]where \(p\) is a positive constant. The convergence of this series depends on the value of \(p\):
  • If \(p > 1\), the p-series converges.
  • If \(p \leq 1\), the p-series diverges.
Understanding the nature of a \(p\)-series is crucial because many series can be directly compared to a \(p\)-series. In the exercise given, the series was compared to the \(p\)-series \(\sum \frac{1}{k^2}\), which is a known convergent series since \(p = 2\) is greater than 1. By this comparison, the convergence of the original series was established.
Rational Function
A rational function is any function that can be expressed as the ratio of two polynomials. In terms of series, a common general term within a series might be a rational function. Identifying a function as rational is helpful, particularly when trying to simplify or factor, as was illustrated in the solution to our exercise. In the given exercise, the term \[\frac{1}{16k^2 + 8k - 3}\]is a rational function, since both the numerator (which is just 1) and the denominator (which is a quadratic polynomial) fit this description. Simplifying or understanding the behavior of rational functions can help us apply tests like the Comparison Test, as these expressions often lend themselves to meaningful comparison with simpler series such as \(p\)-series.
Series Convergence
Series convergence refers to whether the sum of the infinite series approaches a finite number as more terms are added. Determining convergence is fundamental when dealing with series because it tells us if the series "sums up" to a real number or if it goes to infinity. Convergence can be tested using various tests like the Comparison Test, Ratio Test, and others.In the stated exercise, series convergence was determined through using the Comparison Test against a known convergent \(p\)-series. Once a series is concluded to be convergent, it ensures that the summation of its infinite terms is stable and results in a finite sum.Understanding series convergence allows us to evaluate not just the behavior of individual terms but the ultimate behavior and result of the entire series itself. This concept is pivotal in advanced calculus and analysis, playing a role in understanding infinite processes and sums.

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Most popular questions from this chapter

The famous Fibonacci sequence was proposed by Leonardo Pisano, also known as Fibonacci, in about A.D. 1200 as a model for the growth of rabbit populations. It is given by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1\) and \(f_{1}=1 .\) Each term of the sequence is the sum of its two predecessors. a. Write out the first ten terms of the sequence. b. Is the sequence bounded? c. Estimate or determine \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}},\) the ratio of the successive terms of the sequence. Provide evidence that \(\varphi=\frac{1+\sqrt{5}}{2},\) a number known as the golden mean. d. Use induction to verify the remarkable result that \(f_{n}=\frac{1}{\sqrt{5}}\left(\varphi^{n}-(-1)^{n} \varphi^{-n}\right)\)

The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b > 1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n .\) Use a calculator to determine the smallest value of \(n\) such that \(n ! > b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\)

Identify a convergence test for each of the following series. If necessary, explain how to simplify or rewrite the series before applying the convergence test. You do not need to carry out the convergence test. $$\sum_{k=1}^{\infty} \frac{k^{2}}{k^{4}+k^{3}+1}$$

Identify a convergence test for each of the following series. If necessary, explain how to simplify or rewrite the series before applying the convergence test. You do not need to carry out the convergence test. $$\sum_{k=10}^{\infty} \frac{1}{(k-9)^{5}}$$

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$

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