![]() ![]() Using a computer, we find that Newton’s Method seems to converge to a solution x = 9118.01 after 8 iterations. We find f ′ ( x ) = ( 1 - ln ( x ) ) / x 2. X n + 1 = x n - f ( x n ) f ′ ( x n ). Recall how Newton’s Method works: given an approximate solution x n, our next approximation x n + 1 is given by ![]() We make a guess that x must be “large,” so our initial guess will be x 1 = 1000. Let f ( x ) = ln ( x ) / x - 0.001 we want to know where f ( x ) = 0. This cannot be solved algebraically, so we will use Newton’s Method to approximate a solution. We start by solving ( ln n ) / n = 0.001 for n. We want to find n where ( ln n ) / n ≤ 0.001. The important lesson here is that as before, if a series fails to meet the criteria of the Alternating Series Test on only a finite number of terms, we can still apply the test. We can apply the Alternating Series Test to the series when we start with n = 3 and conclude that ∑ n = 3 ∞ ( - 1 ) n ln n n converges adding the terms with n = 2 does not change the convergence (i.e., we apply Theorem 9.2.5). The derivative is negative for all n ≥ 3 (actually, for all n > e), meaning b ( n ) = b n is decreasing on [ 3, ∞ ). Treating b n = b ( n ) as a continuous function of n defined on [ 2, ∞ ), we can take its derivative: Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License. We recommend using aĪuthors: Gilbert Strang, Edwin “Jed” Herman Use the information below to generate a citation. Then you must include on every digital page view the following attribution: If you are redistributing all or part of this book in a digital format, Then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a print format, Want to cite, share, or modify this book? This book uses theĬreative Commons Attribution-NonCommercial-ShareAlike License Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled. The two cards below B B are now in random order. The expected number of top random insertions before this happens is n / 2. Once one card is below B, B, there are two places below B B and the probability that a randomly inserted card will fall below B B is 2 / n. Thus the expected number of top random insertions before B B is no longer at the bottom is n. If the deck has n n cards, then the probability that the insertion will be below the card initially at the bottom (call this card B ) B ) is 1 / n. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. In Figure 5.12, we depict the harmonic series by sketching a sequence of rectangles with areas 1, 1 / 2, 1 / 3, 1 / 4 ,… 1, 1 / 2, 1 / 3, 1 / 4 ,… along with the function f ( x ) = 1 / x. To illustrate how the integral test works, use the harmonic series as an example. It is important to note that this test can only be applied when we are considering a series whose terms are all positive. ![]() ![]() This test, called the integral test, compares an infinite sum to an improper integral. This technique is important because it is used to prove the divergence or convergence of many other series. In this section we use a different technique to prove the divergence of the harmonic series. In the previous section, we determined the convergence or divergence of several series by explicitly calculating the limit of the sequence of partial sums and showing that S 2 k > 1 + k / 2 S 2 k > 1 + k / 2 for all positive integers k. 5.3.3 Estimate the value of a series by finding bounds on its remainder term.5.3.2 Use the integral test to determine the convergence of a series.5.3.1 Use the divergence test to determine whether a series converges or diverges. ![]()
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