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Summer Calculus BC Homework Part 7 Theory of Infinite Series a The expression is an infinite series. It represents the sum a1 a2 a3 i i 1 ai ai 1 The beginning index, in this case i 1 can start at any finite value; there are still an infinite number of terms since the sum continues forever. It is often preferable to write this sum as a n limiting statement, such as lim ai . n i 1 We are most often concerned with whether or not a series is convergent or divergent—that is, whether or not the sum of all the infinitely many terms exists and is itself finite, or not. An infinite series is convergent if its sequence of partial sums is convergent. That is, for the infinite series a let S1 a1 , S2 a1 a2 , S3 a1 a2 a3 , S4 a1 a2 a3 a4 and so on, i i 1 thus Sn a1 a2 a3 an . a i 1 i will exist as a finite value if and only if ( S n ) S . Note that a series does not have to become arbitrarily large for it to fail to converge. For example, the series (1) i is given by 1 1 1 1 1 and is divergent. i 1 What should make intuitive sense is that in order to add an infinite amount of terms together yet attain a finite sum from all of those terms, not only does each term have to be finite, but ultimately the terms each need to approach zero. That is not saying the value of the entire series should be zero, merely that if you add up an infinite amount of things, in the end you have to be adding an infinite amount of zeroes in order to get a finite sum. This idea is given explicitly as the test for divergence: If lim ak 0 , then the series k a k diverges. Notice in the statement above that the starting index is not specified, but it is understood that the series starts at some finite value of k and proceeds to have k approach infinity. For example, k 1 k k 2 diverges, since lim ak lim k k 2 1 0 . Be very careful to not think of this k k test for divergence as being a biconditional. Just because lim 1k 0 does not mean that k k 1 1 k exists (and in fact, it doesn’t—the sum is infinite). An infinite series of the form n 1 n 1 n 1 is called a p–series. In the case where p 1 we have just 1 n 1 12 13 14 from above, which is known as the harmonic series. If p 2 our series is 1 n2 1 14 19 161 and so on. 1 np We will take it axiomatically for now that a p–series converges for p 1 and diverges for p 1 . The proof of this claim is best done using a technique from Calculus called integration, hence we must take the claim for now as a postulate. Nonetheless we are still right now able to show that the harmonic series diverges—consider the partial sums: S1 1 , S2 1 12 , S 4 1 12 13 14 1 12 12 , S8 1 12 13 14 15 18 1 12 12 12 , You should see that the sequence N of partial sums diverges. Note that the harmonic series diverges very slowly—for k 1 need N 1043 . On the other hand, n 1 1 n2 converges and 1 14 91 161 1 k 100 , we 6 . The proof of that 2 claim will have to wait for more advanced courses. Frequently we are more interested in whether or not a series converges than what it may actually converge to. Therefore, making comparisons are often useful in determining if a given infinite series converges or diverges. a a 1 If k k and b are infinite series with 0 ak bk for all k, then if k b k converges, so does We call this the basic comparison test. The basic comparison test should be obvious in an intuitive sense as well: If team A scores less than (or equal to) team B in every inning of a baseball game, then by the end of the game, team A has scored a lower (or at most equal) total than team B. In our case, there are an infinite number of innings and we are saying that if team B has a finite score, then the score for team A must also be finite since it is less than (or at most equal to) team B’s score. 1 1 1 . Since n n for all 2 n 2 n 1 2 n 1 1 n 1 , and n is a convergent geometric series with common ratio , 2 must be a finite real 2 2 n 1 number. It is possible to approximate that value of 2 to any desired number of decimal places, but it is otherwise a transcendental value that cannot be written in terms of other elementary functions. For example, the Cote number of order 2 is defined as 2 n Again, the student here is cautioned to be careful not to read too much into the basic comparison test. If ak converges, that does not necessarily say anything about the convergence or b divergence of a divergence of k . If k . b k diverges, that does not say anything about the convergence or In all of this analysis, we don’t need to start at k 1 either. Let N be a positive integer. The series a k 1 1 k converges if and only if the series a kN k converges for each integer N 1 . We can loosen up the restriction that 0 ak bk for all values of k by realizing that bk can actually be larger than its corresponding a k for any finite amount of times, but there just has to be some point in each series where from then on, 0 ak bk . We say that mathematically as “there exists a value K such that k K implies 0 ak bk .” We don’t know what K is, but as long as it exists, we are good. Interestingly enough, although the harmonic series diverges, the series 1 12 13 14 15 converges. We call that series the alternating harmonic series, and as will be seen in Calculus, it converges to ln 2 . In general an alternating series is of the form (1) k k 0 ak a0 a1 a2 a3 a4 . We take each ak as being non-negative here. Again axiomatically, alternating series converge when (i) the terms of the sequence (ak ) are decreasing and (ii) lim (ak ) 0 . We call this the alternating series test. One of the more useful results for convergent alternating series is that using a partial sum to estimate the true value of the entire series will be in error by no greater than the absolute value of the first truncated term, an 1 . The diagram below illustrates this: a2 a4 S a3 a1 S1 S2 S3 S4 S5 S6 S7 S8 For a1 0 , the red line in the diagram shows how the error in approximating S by S1 is less than a2 . Likewise, the blue line in the diagram shows how the error in approximating S by S 2 is less than a3 . The purple line shows how the error in approximating S by S3 is less than a4 , etc. Starting in second year algebra you worked with infinite series that were geometric, and you learned a few useful results for them. Consider a n . This is really an infinite geometric series n 0 1 a a 2 a3 a 4 whose common ratio is a. a1 where a1 is 1 r the first term of the series (in this case it is 1) and r is the common ratio for the series (in this case it is a). The sum S will exist when r 1 . Do not overlook the importance of that last requirement for the sum S to exist. If a geometric series is convergent, then the sum of the series is given by S Consider the infinite geometric series 1 32 94 278 Since here, r 32 , the sum increases without bound and the series diverges. However, 4 2 1 12 18 converges to a sum of 8. is convergent and in fact, it Take the infinite geometric series 1 13 ( x 2) 19 ( x 2) 2 271 ( x 2)3 . The series may converge or diverge, depending upon the value of the common ratio. Here the common ratio is 1 ( x 2) and so needing r 1 implies that 13 ( x 2) 1 which is equivalent to 5 x 1 . 3 Hence, the series 1 13 ( x 2) 19 ( x 2) 2 271 ( x 2)3 diverges otherwise. For an arbitrary infinite series a n converges only when 5 x 1 and we have one last result for now which we call the ratio test. If the ratio of consecutive terms, in particular ak 1 , consistently approaches a constant, then the ak series is basically acting like it is a geometric series. In other words, if lim n an 1 (here we are an using the Greek letter rho, rather than r, for common ratio) then instead of needing r 1 we would then need 1 . We have to be careful— if 1 the series will diverge but if 1 we cannot draw a conclusion from the ratio test and need to apply another test for convergence. The series k 1 k3 3k is not itself geometric, but since here ak 3kk and ak 1 ( k3k 1)1 , we have 3 3 ( n 1)3 a lim n 1 lim n a n n 3n1 n3 3n lim 1 n 3 ( nn31) 13 and since 3 1 3 1, k 1 k3 3k must be convergent. We don’t know what the sum of the series is, but we know the sum must exist. Out there. Somewhere. (1)k 1 x k 1 an 1 The series lim is not geometric either. However, lim k n n an k 2 k 1 ( 1)n2 x n ( n 1)2n1 ( 1)n1 x n1 n 2 . n Lots of algebra happens, and we can drop all of the ( 1) stuff since we are taking absolute n values and get lim n xn n 2n x n x n x x lim lim 1 . n 1 n 1 n 2 n 1 n 1 2 2 n n 1 2 2 x x 1 , so x 2 –e.g. the series definitely 2 converges for 2 x 2 . But we have to check on the case where 1 separately. If x 2 To converge we need 1 which implies we need which is a divergent harmonic series. But when x 2 , the series is which is an alternating series and does converge. Thus, overall the then the series is k 1 ( 1)k 1 ( 2)k 1 k 2 k ( 1)k 1 ( 1)k 1 2k 1 k 1 k 2 k k 1 k 2 k 1 k 1 ( 1)k 1 21 k k 1 ( 1) k 1 2k original series converges for 2 x 2 . ( 1)2 k 2k 21 k k 1 ( 1)k 1 (2)k 1 k 2k 21 k k 1 k 1 1 2k ( 1)k 1 2k 21 k 2 k Name: Date: Summer Calculus Homework Part 7A Theory of Infinite Series Period: Seat: 1. Consider the infinite geometric sequence a1 , a2 , a3 , a4 , a5 , a6 , … with common ratio r. If a an is the nth term of the sequence, which of the following is equal to n 1 ? an1 (A) 1 r (B) 2 r (D) r 2 (C) 2r (E) 1 r2 an 0 n 2. The graph above shows the terms of a sequence. Which of the following statements best describes the sequence? (A) The sequence is divergent (B) The sequence can only be described by a recursive relationship (C) The sequence is geometric with common ratio r, where 1 r 0 (D) The sequence is geometric with common ratio r, where 0 r 1 (E) The sequence is geometric with common ratio r, where r 1 3. Which of the following expressions is equivalent to 1 2 x 4 x2 8x3 16 x 4 (A) 2 x n (B) n 1 (D) (1) n (2 x)n 1 (C) n 1 2xn (E) n 1 2 x ( 2 x ) n0 n 1 n 0 4. Which of the following infinite series is divergent? (A) 4 ( 23 )n (B) n 1 n 1 (D) ( 4n 1 n2 ) n 1 4 ( 23 )n (E) e n0 n (C) ( n 1 2 n 1 8 n 1 ) n 1 5. Which of the following infinite series is divergent? (A) ( 76 )n (B) n 1 (D) n0 ( 3n1 n ) (C) n 1 ( 1) ln( n 2) n (E) n 1 n 1 1 n 2 n (2 n )! 6. Which of the following gives all values of x such that is a convergent geometric series? xn 5n n 1 (A) x 5 (D) 5 x 5 (B) 0 x 5 (E) 15 x 15 7. Consider the infinite geometric series 1 (2 x 1) (2 x 1) 2 (2 x 1)3 . Give an inequality of the form a x b for x to ensure that this series converges. (C) 0 x 5 8. Give an infinite geometric series whose sum equals 5. Hint: Find the value of x such that 1 x x2 x3 x 4 5 9. Compute each sum or explain why the sum does not exist. Give simplified and exact values for sums, not decimal approximations. A. n 0 3 n 4 C. sin 1 2 k 0 B. tan 2 2 k k 0 k D. 3 3 3 3 9 2 4 8 16 Name: Date: Summer Calculus Homework Part 7B Theory of Infinite Series Period: Seat: 1. Let S be the sum of an infinite geometric series. For each of the following, write a series whose sum is: A. S 1 1 2x B. S 1 1 13 x C. S 3 4 x2 (hint: multiply and divide by 14 ) 2. Perform long division on 1 (1 x) to obtain an infinite termed polynomial. This polynomial is only valid for x 1 . Use your knowledge of infinite geometric series to explain why this is so. (1 x) 1 3. Verify that the infinite geometric series is convergent. Use the series to demonstrate how 1 4n n 1 the basic comparison test can be used to show each of the following series are convergent. A. n 1 1 5n B. 1 4 n n n 1 C. n 0 1 n! 4. What is a reasonable upper bound on the error of estimating the value of the series k 1 ( 1) k k 2k by the partial sum S4 12 12 83 14 ? Is S 4 an under approximation or an over approximation of the value of the series? For problems 5 and 6 below, A. Set up a limit lim n an1 an , B. Use your limit to calculate the value of so you can, C. Conclude whether the infinite series converges or diverges. n 5. j 0 j ( j31 ) 3n n 1 n ! 6.