Assignment 6 Assigned Weds Oct Section 8, Problem 3 a, a 3, a 3 5, a 4 7 Section 8, Problem 4 a, a 3, a 3, a 4 3 Section 8, Problem 9 a, a, a 3, a 4 4, a 5 8, a 6 6, a 7 3, a 8 64, a 9 8, a 0 56 Section 8, Problem This is the famous Fibonacci sequence Each term is the sum of the previous two terms a, a, a 3, a 4 3, a 5 5, a 6 8, a 7 3, a 8, a 9 34, a 0 55 Section 8, Problem 6 We want the signs to alternate, so use ) n+, and the denominators are the squares, so a n )n+ n Section 8, Problem 0 We want to start with and then we want to add on 4 for each new term So when n we add 0, when n we add 4, when n 3 we add 8, and so on This means we need to add 4n ) to This gives the formula a n + 4n ) 4n 54
Section 8, Problem 4 This sequence converges, lim a n + ) n n lim lim n n n + )n n n Section 8, Problem 7 This sequence also converges, lim a n lim n n 5n 4 n 4 + 8n 3 lim n n 4 5 5 + 8n Section 8, Problem 3 This sequence diverges, since it looks like 0,, 0,, 0,, 0,, 0,, Since it flips back and forth between 0 and, it does not converge to any one number Section 8, Problem 4 This one converges, since the numerator is never larger than, while the denominator goes to infinity, so lim a sin n n lim n n n 0 55
Assignment 7 Assigned Fri Oct 5 Section 8, Problem 59 Here a n looks like a n n! n n ) n ) 3 nn n n n n n n n n 3 n n n All of the fractions in the product are less than or equal to, and half of them the ones the look like k n with k n are less than or equal to So the product includes at least n/ factors that are less than or equal to Therefore a n ) n/ The quantity /) n/ goes to 0 as n goes to, so lim a n 0 n Section 8, Problem 6 Again we write a n as a product a n n! 0 6n n n ) n ) 3 0 6n n 0 6 n 0 6 n 0 6 3 0 6 0 6 0 6 So as soon as n gets very large, each new a n is much larger than the previous one To be precise, if n > 0 6, then a n is at least as big as a n, so eventually the sequence of a n s more than doubles with each successive term Therefore the limit does not exist 56
Section 8, Problem 68 This one is not so clear Let y x ) x What happens to y when x? This is one of those indeterminate forms We first simplify by taking logarithms, ln y x ln x ) ln ) x We compute the limit using L Hôpital s rule ln ) x x lim ln y lim x x lim x x x 0 This is the limit of the logarithm of y, so lim y lim ) x x x x e 0 x Hence lim a n lim ) n n n n Section 8, Problem 70 Let s simplify the formula by multiplying the numerator and the denominator by 0 ) n This gives a n 0/) n 0 )n 9/0) n + /) n 0 ) n 00 n 88 n + 0 n This goes to 0 because the 0 n will dominate To be precise, we can write it as lim a n lim n n 00 n 88 n + 0 n lim n 00/0) n 88/0) n + 0 0 Section 8, Problem 85 x, x, x 3, x 4 4, x 5 8, x 6 6, x 7 3, So x n is a power of, and adjusting to get the right power of two, we get x and x n n for n 57
Section 8, Problem 97 We can rewrite the formula for a n as a n 3n + n + 3n + 3 3n + 3 n + n + n + 3 n + So the sequence is nondecreasing in fact, it is strictly increasing), since as n gets bigger, we re subtracting less from 3 The sequence is bounded above by 3 Section 8, Problem 98 We can cancel the factors in the denominator from the factors in the numerator, so n + 3)! a n n + 3)n + )n + ) n + 3)n + ) n + )! Thus a n is nondecreasing in fact, it is strictly increasing) It is not bounded above, since for example a n is larger than n + ) n+, since it s a product of n + terms, each of which is at least as big as n + Section 8, Problem 5 No a sequence of positive numbers that is bounded above might diverge because it might never settle down to be close to a single value Here s a simple example from an earlier homework, a n + ) n a n flips back and forth between and 3, so it doesn t converge, but the values are positive and bounded Section 8, Problem This is a geometric series with initial term a and common ratio r 3, so s n + 3 + 9 + + 3 n 3n 3 3 ) 3 n Then lim s n lim 3 ) n n 3 n 3 58
Section 8, Problem 4 We have s, s, s 3 3, s 4 5, s 5, This is also a geometric series with initial term a and common ratio r So s n + 4 8 + + The limit does not exist )n ) n ) ) n ) 3 59
Section 8, Problem 5 We compute the first few values s 6, s 4, s 3 3 0, s 4 3, s 5 5 4, s 6 3 8, s 7 7 8, s 8 5, s 9 9, s 0 5, s 6 The pattern may not be obvious, but try looking just at the values of s n when n is odd These are s 6, s 3 3 0, s 5 5 4, s 7 7 8, s 9 9, s 6 Now the pattern is clear at least for odd n), namely s n n n + 4 And you can go back and check that this works for the even values of n, too The reason it was confusing is because there was some cancelation in the fractions for even n) Then the limit is lim s n n lim n n n + 4 If you want to check that the formula for s n is correct for all n, you can write n + )n + ) n + n + Then s n 3 + 3 4 + 4 5 + + n + )n + ) ) + 3 3 ) + 4 4 ) + + 5 n + ) n + Notice how the 3 terms cancels, and the 4 terms cancel, etc So at the end, all you re left with is the first term, which is, and the last term, which is n+ This is called a telescoping sum) So the partial sum equals s n n + n n + ) 60