Advanced Calculus II Unit 7.3: 7.3.1a, 7.3.3a, 7.3.6b, 7.3.6f, 7.3.6h Unit 7.4: 7.4.1b, 7.4.1c, 7.4.2b, 7.4.3, 7.4.6, 7.4.7

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1 Advanced Calculus II Unit 73: 73a, 733a, 736b, 736f, 736h Unit 74: 74b, 74c, 74b, 743, 746, 747 Megan Bryant October 9, 03 73a Prove the following: If lim p a = A, for some p >, then a converges absolutely We now that lim p a = A, p > implies that lim p a = A N N such that for all > N, p a A < Thus, a < A + p = ( A + ) p, where ( A + ) is a number We now that for p >, p is a convergent p-series by the comparison test, we now that a converges and a converges absolutely 733a Prove that if a converges and b converges absolutely, then a b converges Since a converges, lim a = 0 and a is bounded for some a M, N for some M R Since b converges absolutely, by the Cauchy Criterion (theorem 73), give ɛ > 0, N N such that for all n > m > N,

2 S n S m = a n a n + + a m+ ɛ M, where S n is the n th partial sum of a For some N N, a n b n a m b m = a n b n + a n b n + + a m+ b m+ a n b n + a n b n + + a m+ b m+ M( a n + a n + + a m+ M( a n + a n + + a m+ < M ɛ M = ɛ if a converges and b converges absolutely, a b converges 736b Test the following series for absolute and conditional convergence: =3 ( ) ln(ln ) First, we will apply the alternating series test Let b = ln(ln ) Let f(x) = x ln(ln x) Thus, f (x) = decreasing lim x x ln(ln x) = 0 ( ln(x) ln(ln(x))) (x ( 3/) ln(x) ln (ln(x))) < 0, x [3, ) Thus, b is monotone Thus, by the alternating series test, we now that Next, we will test b = ln(ln ) for convergence We will use the comparison test with a = n ln(ln ) > ln > 3/ >, [3, ) =3 ( ) ln(ln ) converges by the comparison test, we now that b = ln(ln ) diverges for all [3, ) ( ) ln(ln ) converges conditionally =3

3 736f Test the following series for absolute and conditional convergence: ( ) + (+) + First, we will use the alternating series test: Let b = Let f(x) = (+) + x x (x+) x+ Then f (x) = (xx )(ln(x) ln(x+)) < 0, x [, ) b (x+) x+ is monotone decreasing lim (+) + (+) (+) = 0 by the alternating series test, Next, we will evaluate ( ) + (+) + = We will use the comparison test with a = lim (+) + + (+) + = e ( ) + (+) + (+) + 0 < e <, therefore by the limit comparison test, 736h ( ) + (+) + converges conditionally converges for convergence (+) + Test the following series for absolute and conditional convergence: ( ) + sin( ) We will use the alternating series test: Let b = sin( ) diverges Since 0 π for N and sin is monotonically decreasing over [ π, 0], we now that b is monotonically decreasing 3

4 lim b sin( = sin( lim ) = sin(0) = 0 by the alternating series test, ( ) + sin( ) < ( ) + sin( ) = sin( ) We will use the limit comparison test: Let a = and b = sin( ) b lim a sin( sin() = Since, 0 < <, we now that b converges if and only if a coverges a is the harmonic series, which is divergent Thus, b = sin( ) diverges Therefore ( ) + sin( ) converges conditionally 74b Determine if the following sequence is in l : { ln } = Let a = ln a = a = = ( = ln ) = = ln () We will use the integral test to test for convergence: Let f(x) = x ln (x) f (x) = ln(x)+ < 0 for all x [, ) x ln 3 (x) f(x) is monotone decreasing f(x)dx = x ln (x) n ln x = 0 < = ( ln ) converges and the sequence { ln } = l 4

5 74c Determine if the following sequence is in l : { ln } = Let a = ln a = a = = ( ln() = ) = = ln () We will use the comparison test to test for convergence: Let b =, which is the harmonic series and, thus, divergent We now that 0 ln () for > e since ln () > when > e by the comparison test, since = diverges, ln () diverges = Thus, the sequence { ln } = is not in l 74b Determine all values of p R such that the given sequence is in l : { p p } Let a = p p a = ( p p ) = p p We will first use the root test: lim sup p p sup p p = (lim sup( / )) p = = p p p by the root test, we now that converges if p > ( p Now, we must examine the case when p = If p =, a = =, which we now diverges ( p p ) converges if and only if p > p ) diverges if p < and 5

6 Thus the sequence { p p } is in l if and only if p > 743 If {a } l, prove that lim a = 0 {a } l implies that a = a < If a <, a < a < implies lim a = 0, by the definition of convergence Which implies that lim a = 0 and lim a = a)suppose that {p } is a sequence in Rn, where for each N, p = (p,,, p,n ) Prove that the sequence {p } converges to p = (p,, p n ) in the norm if and only if lim p,i = p i for all i =,, n Assume that lim p = p Given ɛ > 0, there exists N N such that for all n > N, so that p n = (p p, ) + + (p n, ) < ɛ For a fixed i, p i p i, = (a i a i,n ) p n = (p p, ) + + (p n, ) < ɛ Thus, for each ɛ > 0, lim n p i, = p i Now, for the other direction, let lim n p i, = p i Given ɛ > 0, for each i, let ɛ i = ɛ Let N = max{n,, N n } ɛ n and choose N i N such that p i p i, < Then, for every i [0, n], we have (p i p i, ) < ɛ when > N Therefore p = (a a, ) + + (a n a n, ) < ɛ + + ɛ < nɛ = ɛ Thus the sequence {p } converges to p = (p,, p n ) in the norm if and only if lim p,i = p i for all i =,, n 6

7 b)prove that every bounded sequence in R n has a convergent subsequence Let {a } be a bounded sequence in R n Since {x } is bounded, there exists a, b R n such that x [a, b ], N Let I = [a, b ] such that I + I and (b + a + = b a = = b a 0 Then, by the Nested Interval Theorem, there exists some x 0 such that x 0 is in the intersection of all I, N We will construct a subsequence by choosing x l [a, b ], x l [a, b ] such that l > l for all N By design, we have a x l b with a 0 and b 0 Thus, by the squeeze theorem, x l 0 every bounded sequence in R n has a convergent subsequence 747 For each n N, let e n be the sequence in l defined by { 0, if n e n () =, if = n Show that e n e m = if n m (Remar: The sequence {e n } is a bounded sequence l with no convergent subsequence Thus the Bolzano- Weierstrass theorem (4) fails in l We now that for n N, there is only one instance in which n = m, so en = and e m = In addition, for the e n e m, since n and m are different, we now the sum will always be 0 e n e m = (en e m ) = e n e n e m + e m = + = 7