Solutions to Homework 7

Transcription

1 Solutios to Homework 7 Due Wedesday, August 4, 004. Chapter 4.1) 3, 4, 9, 0, 7, 30. Chapter 4.) 4, 9, 10, 11, 1. Chapter 4.1. Solutio to problem 3. The sum has the form a 1 a + a 3 with a k = 1/k. Sice the a k are positive ad decreasig, the series coverges by the alteratig series test. Let L be the sum of the series. The proof of the alteratig series test shows that S k 1 L S k ad S k S k 1 = a k. Sice the sum of the series is betwee 0.1 ad 1, to guaratee digit accuracy we must have a k < (so we ca be sure whether to roud the th digit up or dow), i.e., 10 < k. For 10 digit accuracy: < k. For 0 digit accuracy: 10 0 < k. For 100 digit accuracy: < k. (The poit is the series coverges very slowly). Solutio to problem 4. It is possible to have differet A s for each ǫ (for example the Cauchy criteria does this), but if a series does ot coverge the it must diverge: if you have a differet value of A for each ǫ, the the sequece of A s is a Cauchy sequece which coverge to some A which works for all ǫ. I.e., let ǫ N be a sequece tedig to 0 + ad A N be a sequece such that A N E < ǫ N for all N. The if N < M we have for all M, A M A N A M E + A N E ǫ N + ǫ M. Sice both ǫ M ad ǫ N go to zero we ca make this less tha ay fixed ǫ by requirig N to be sufficietly large. Thus the A N are a Cauchy sequece ad coverge to some A. The the series coverges to this A: give ǫ > 0 chose N such that A A N < ǫ/ ad ǫ N < ǫ/. The for all N, A E A A N + A N E ǫ/ + ǫ/. 1

2 Solutio to problem 9. The ratio test gives B k+1 B k (k 1)(k)10 k 1 (k + 1)(k + )10 k+1 The factors ot ivolvig the B k approach 10 as k. The after doig some algebra the factors ivolvig B k become B k+1 B k = (k + ) (1 + 1/k) k 1 π e k which goes to ifiity as k. So the series does ot coverge. Solutio to problem 0. Choose N such that for all N, a +1 /a α. (We may assume α > 0, sice if α = 0 the a N+1 = 0 ad the ratio a N+ /a N+1 is ot defied.) We will prove by iductio that for all N, a a N α N. We have clearly that a N+1 α a N. Fix a > N + 1 ad assume that a a N α N. The a +1 α a α a N α N = a N α +1 N, which proves the result by iductio. As a immediate cosequece we have a α( a N α N ) 1/. Now ( a N α N ) 1/ = e 1 l( an α N) = 1. So give ǫ > 0 we ca choose N 1 such that for all N 1, ad i particular, ( a N α N ) 1/ 1 < ǫ/α ( a N α N ) 1/ < 1 + ǫ/α. The for all M = max(n, N 1 ) we have a α( a N α N ) 1/. < α(1 + ǫ/α) = α + ǫ. To show that it does ot ecessarily imply a α, cosider the (diverget) series +++. The a +1 /a = 1, but a = 1/ > 1. (But of course a = 1.) Solutio to problem 7. S 10 = S 100 = S 300 =

3 We have k ( k ) k k = 1 k 1 so the series coverges by the root test. (Note oe would expect the series to coverge also by compariso with a geometric series, sice for ay α (1/, 1), all but fiitely may of the terms are less tha α.) Solutio to problem 30. S 10 = S 100 = S 300 = Oe approach to prove divergece would be to use the it ratio test: a k+1 k a k (k + 1) k+1 = k! k (k + 1)! k k = k k Sice e > 1 the series diverges. Chapter 4.. Solutio to problem 4. 1 (k + 1) k + 1 ( k) k = e. Here is a graph of S (x) for = 3, 6, 9, ad 1: 3 1 ( k + 1 k ) k The graphs seem to approximate each other very well for x less tha about.3 but the they seem to diverge from each other whe x is larger 3

4 tha about.3. A easy computatio shows the radius of covergece is 1/e = So oe expects these graphs to approximate the fuctio well o the iterval x < 1/e. Solutio to problem 9. f(x) = k=1 k k k! xk By the ratio test we fid the radius of covergece is 1. Solutio to problem 10. Let a k = 4 6 k (k + 1) Oe ca use Gauss s test from sectio 4.3. Sice this is sectio 4., I ll try to write a proof without usig it. I ll eed a series of lemmas: Lemma 1. For all k, a k k+1. Proof. For k = 1, a 1 = 3 3. Suppose a k k+1. The sice k+ k+1 > 1. a k+1 = a k k + k + 3 k + 1 k + k + 3 k + 3 Lemma. For all k, a k 1 k. Proof. Use Lemma 1 ad the easily verified fact that k + 1 = 1 k + 1/ 1 k. Corollary. The series diverges at x = 1 by compariso with the harmoic series. Lemma 3. For all k, a k a k+1 0. Proof. Note a k+1 = a k k+ k+3 greater tha or equal to 0. Lemma 4. k 0 a k = 0. Proof. Note the a k ca be writte as a k = k k! (k+1)! k k! ad ote k+ k+3 1. Clearly all the a k s are = k (k!) (k + 1)!. 4

5 Stirlig s formula says that k! = ( ) k k πk e E(k) e where k E(k) = 0 (equatio (A.77), page 301). Pluggig this ito the above ad simplifyig a bit gives a k = ( ) k k e k k + 1 π(k + 1) k + 1 ee(k) E(k+1). As k, the first factor approaches the it e 1 ad the last factor approaches 1. The it s easy to see that a k goes to zero as k. Corollary. The series coverges at x = 1. Proof. The series whe x = 1 is ( 1) k a k k=1 which coverges by Lemmas 3 ad 4 ad the alteratig series test. Solutio to problem 11. Usig the hit 1 3 ( 1) = ()!! ad Stirlig s formula (equatio (A.77), page 301) gives 1 3 ( 1) = ()!! = 1 ( e( e ) π e E() ) π e E() = 1 ( e e ) e E() E() ( = + 1 ) e E() E() e = + 1 e +F() where F() = E() E() goes to zero as (sice E() ad E() do). If we igore the F(), the the ratio of the two sides of the equatio (4.11) gets close to 1, but the absolute value of the differece betwee them gets large. If we look at the ratio of both sides, they are approximately , , ad respectively whe = 10, 0, ad 100 (the ratios are very close to oe). 5

6 If we look at the absolute value of the differece betwee both sides, they are approximately , , ad whe = 10, 0, ad 100 respectively. So without the error terms it is oly i the first sese that the two are good approximatios of each other. Solutio to problem 1. Usig ad Stirlig s Formula gives ( 1) = ()!! so the radius of covergece is e ( 1) = e, 6