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Use the test of your choice to determine whether the following series converge. $$\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\dots$$

Short Answer

Expert verified
Answer: The given series converges.

Step by step solution

01

Identify the given series and the simpler series

The given series is: $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}$$ We will compare it to the simpler converging series: $$\sum_{n=1}^{\infty}\frac{1}{n^2}$$
02

Perform the Limit Comparison Test

We will now perform the limit comparison test by taking the limit of the ratio of the given series and the simpler series as n goes to infinity: $$\lim_{n\to\infty} \frac{\frac{1}{(2n-1)(2n+1)}}{\frac{1}{n^2}}$$
03

Simplify the limit expression

Now we will simplify the expression inside the limit: $$\lim_{n\to\infty} \frac{n^2}{(2n-1)(2n+1)}$$
04

Evaluate the limit

Next, we will evaluate the limit as n goes to infinity: $$\lim_{n\to\infty} \frac{n^2}{(2n-1)(2n+1)} = \lim_{n\to\infty} \frac{n^2}{4n^2-1}$$ Using L'Hôpital's Rule, we get: $$\lim_{n\to\infty} \frac{2n}{8n} = \frac{1}{4}$$ Since the limit is a finite positive number, the given series converges or diverges the same way as the chosen simpler series.
05

State the conclusion

Since the simpler series \(\sum_{n=1}^{\infty}\frac{1}{n^2}\) converges and the limit comparison test result was a finite positive number, this implies that the given series \(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\) also converges.

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Key Concepts

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

Convergent Series
A convergent series is a sequence of numbers that approaches a finite value as more and more terms are added. This concept is fundamental to understanding how infinite series behave. In essence, if the series converges, the sum of its infinite terms will yield a specific, finite value.

For instance, a well-known example of a convergent series is the p-series, where

ewlineewline\[ ewlineewlineewlineewlineewlineewline\sum_{n=1}^{ewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline(2n-1)(2n+1)} ewlineewlineewlineewlineewlineewlineewlineewline\right) = \frac{1}{n^p} ewlineewlineewlineewlineewlineewlineewlineewline\] ewlineewline

Convergence is determined by whether 'p' is greater than 1. This series converges when 'p' is greater than 1 because as 'n' becomes very large, the terms become very small fast enough such that their sum remains bounded. Similarly, if we examine the series presented in the exercise, we use the limit comparison test with one of these known convergent series to determine the behavior of our series of interest.

When we encounter a series whose convergence is not easily determined by a simple rule or pattern, we can compare it to a series that we already know to be convergent or divergent. If the series of interest behaves similarly to the known series (via the limit comparison test), we can deduce its convergence or divergence.
Infinity Series
An infinity series, or simply an infinite series, is a sum of infinitely many terms, and it can present itself in various complexities. When working with an infinite series, it's crucial to understand whether the series converges or diverges. That means we want to know if adding an infinite number of terms will approach a finite limit, or if the sum will be unbounded.

To examine the behavior of these series, mathematicians have developed multiple tests. One such test is the Limit Comparison Test, which is particularly handy when the series does not fit neatly into one of the more straightforward categories of convergence tests, such as the p-series or geometric series tests.

Let's highlight an example within the context of this exercise. We are dealing with an infinite series where the terms are given by the formula:

\[\frac{1}{(2n-1)(2n+1)}\]

Here, we look to compare it to a simpler series that we understand well, such as the series with terms

\[\frac{1}{n^2}\],

which is a known convergent series. Through comparison, we can then extrapolate the behavior of the more complex infinite series in question.
L'Hôpital's Rule
When dealing with limits, especially those that result in indeterminate forms like \[0/0\] or \[ewlineewlineewlineewlineewlineewline(ewlineewlineewlineewlineewlineewline\infty/ewlineewlineewlineewlineewlineewline\infinity)ewlineewline\], L'Hôpital's Rule proves to be an invaluable tool. This rule states that under certain conditions, the limit of a ratio of two functions can be found by taking the limit of the ratio of their derivatives.

To employ L'Hôpital's Rule, one must first encounter a limit that evaluates to a 0/0 or infinite/infinite indeterminate form. If the original functions are differentiable and the limit of their derivatives exists, the rule can be applied.

In the solution to the exercise provided, L'Hôpital's Rule is used in the final step of evaluating the limit of the ratio of the terms from the given series and the comparison series. The rule simplifies the processes of determining limits that would otherwise require more complicated algebraic manipulations or intuitive leaps. Once applied, as in the exercise, it provides us with a clear and manageable pathway to understand that the series under consideration is indeed convergent because the limit is a finite number.

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Most popular questions from this chapter

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}} .\) When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots$$ Use the estimation techniques described in the text to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

\(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \operatorname{In} 1734,\) Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} .\) An elegant proof is outlined here that uses the inequality $$\cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\left(\text { provided that } 0

In Section \(8.3,\) we established that the geometric series \(\sum r^{k}\) converges provided \(|r| < 1\). Notice that if \(-1 < r<0,\) the geometric series is also an alternating series. Use the Alternating Series Test to show that for \(-1 < r <0\), the series \(\sum r^{k}\) converges.

Repeated square roots Consider the expression \(\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},\) where the process continues indefinitely. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=\sqrt{1+a_{n}}\), for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n},\) provided the limit exists. b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\) c. Estimate the limit of the sequence. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. e. Repeat the preceding analysis for the expression \(\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}},}\) where \(p>0 .\) Make a table showing the approximate value of this expression for various values of \(p .\) Does the expression seem to have a limit for all positive values of \(p ?\)

The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

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