Math 320, Real Analysis I
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Exercise 2.5.2. Associativity of Infinite Series: (a) Prove Prove that if an infinite series converges converges,, then the associative property property holds. Proof. Let
a
be an infinite series that converges to L. This means the corresponding sequence of partial sums, (s ( sm ), converges to L. Now consider a regrouping of the terms n
(a1 + a2 + · · · + an1 ) + (a ( an1 +1 + · · · + an2 ) + (a ( an2 +1 + · · · + an3 ) + · · ·
This yields the series bk , where b1 = a1 + a2 + · · · + an1 , b2 = an1 +1 + · · · + an2 , b3 = an2 +1 + · · · + an3 , and in general bk = (an −1 +1 + · · · + an ) for each k ∈ N. We claim that the series bk converges to L as well. So consider consider its sequen sequence ce of partial partial sums, (t (t ), where t = b1 + b2 + · · · + b k
k
for each ∈ N. Now t = b1 + b2 + · · · + b = (a1 + a2 + · · · + an1 ) + (a ( an1+1 + · · · + an2 ) + · · · + (a (an −1 +1 + · · · + an )
= a1 + a2 + · · · + an = sn ,
where sn is the n -th partial sum of the original series an . Therefore, Therefore, the sequence, sequence, (t ), of partial sums of the series bk is a subsequence of the sequence, (s ( sm ), of partial sums of the series an , namely (t (t ) = (sn ). He Henc nce, e, by Theo Theore rem m 2.5 2.5.2 .2,, sinc sincee the the sequence (s (sm ) → L, its subsequence is convergent with the same limit, (s (sn ) → L. That is, (t (t ) → L, so we conclude bk = L, which says that the series obtained by regrouping also converges to L.
(b) Compar Comparee this result result to the example example discus discussed sed at the end of Section Section 2.1 where where infinite infinite additio addition n was shown not to be associa associativ tive. e. Why Why doesn’t our proof in (a) apply to this example? (−1)n = −1 + Solution: The example at the end of Section 2.1 refers to the series 1 + −1 + 1 + −1 + 1 + · · · . The first grouping mentioned is
(−1 + 1) + (−1 + 1) + (−1 + 1) + (−1 + 1) + · · · = 0 + 0 + 0 + 0 + · · · = 0, whereas in the second grouping we have
−1 + (1 + −1) + (1 + −1) + (1 + −1) + · · · = −1 + 0 + 0 + 0 + 0 + · · · = −1. That first grouping corresponds to the subsequence of partial sums (s ( s2n ) = (s2 , s4 , s6 , s8 , . . . ) = (0, (0, 0, 0, 0, . . . ), which clearly converges to zero, while the second grouping corresponds to the subsequence (s (s2n−1 ) = (s1 , s3 , s5 , s7 , . . . ) = (−1, −1, −1, −1, . . . ), which we see converges to −1. Therefore, the sequence of partial sums (s ( sm ) of the series (−1)n has subsequences which converge to different limits. This implies the sequence (s ( sm ) diverges , as in Example 2.2.7. In fact, this is the reason that our proof in (a) does not apply to this example. Since the sequence of partial sums, (s (sm), diverges, the series (−1)n is not convergen convergent. t. Hence Hence n (−1) does not satisfy the hypothesis in Part (a), so our proof does not apply to it.
Math 320, Real Analysis I
Solutions to Homework 3 Problems
Exercise 2.5.6. Let (an ) be a bounded sequence, and define the set S = {x ∈ R : x < an for infinitely many terms an }. Show that there exists a subsequence (an ) converging to s = sup S . k
Proof. Let (an ) be a bounded sequence. Hence, by definition, there is a real number M > 0
such that |an | ≤ M for all n ∈ N. Therefore, −M ≤ an ≤ M for all n. This implies −M − 1 ∈ S , so S is nonempty. Furthermore, if x ∈ S , then there must exist a term an for which x < an . Thus x < an ≤ M , so M is an upper bound for the set S = {x ∈ R : x < an for infinitely many terms an }, so S is bounded above. From this, we conclude that the set S has a supremum, s = sup S , by the Axiom of Completeness . We first show that, for all ε > 0 and for all ∈ N, there are infinitely many terms ak ∈ V ε (s) ε with k ≥ . First, there may only be finitely many an with an > s + , for if there were 2 ε infinitely many of them, then s + ∈ S so that s = sup S . Therefore, there is some M ∈ N 2 ε such that an ≤ s + < s + ε whenever n ≥ M , i.e., there are at most finitely many terms 2 from the sequence that are greater than or equal to s + ε. Then, by Lemma 1.3.7, there must be some element x ∈ S such that s − ε < x. Since x ∈ S , there are infinitely many an such that x < an . Therefore, as s − ε < x, there are infinitely many terms an for which s − ε < an . Of these, only finitely many may be larger than s + ε and only finitely many have indices less than , so there must be infinitely many terms an of the sequence satisfying both an ∈ V ε (s) and n ≥ . Now consider the interval (s − 1, s + 1). By our previous paragraph, there are infinitely many terms an such that an ∈ (s − 1, s + 1) and n ≥ 1. Let n1 be the smallest index such that the term an1 ∈ (s − 1, s + 1). Then we consider the smaller interval (s − 12 , s + 12 ), which must contain infinitely many terms an such that n ≥ n1 + 1. Let n2 be the smallest such index for which n2 ≥ n1 + 1 > n1 and an2 ∈ (s − 12 , s + 12 ). Continue in this way, selecting nk to be the smallest index such that nk ≥ nk−1 + 1 > nk−1 and an ∈ (s − k1 , s + k1 ), which exists since there are still infinitely many an ’s with n ≥ nk−1 + 1 and an ∈ (s − k1 , s + k1 ) by the previous paragraph. Then we have inductively constructed a subsequence (an ) of our given sequence, (an ), which we now claim converges to s. k
k
Let ε > 0 be arbitrary. By the Archimedean Property, there is a natural number K such that 1 1 1 < ε. Suppose k ≥ K . Then ≤ < ε, and K k K an ∈ k
1 1 s− ,s+ k k
,
so
|an − s| < k
1 <ε k
whenever k ≥ K . Therefore, (an ) → s. k
Exercise 2.6.6. Equivalent Statements of the Completeness of the Real Numbers: (a) Use the Nested Interval Property to give a proof of the Axiom of Completeness. Proof. Suppose that A ⊆ R is nonempty and bounded above. We want to show that A
has a least upper bound, which is the conclusion of the Axiom of Completeness. Since A is nonempty, there is some element a ∈ A. As A is bounded above, there is a real
Math 320, Real Analysis I
Solutions to Homework 3 Problems
number b ∈ R such that a ≤ b for all a ∈ A. Now consider the interval I 1 = [a, b], which contains an element of A and an upper bound for A. If a = b, then a is an upper bound for A and, if b is any other upper bound of A, then a ≤ b (since a ∈ A), so a = sup A by definition of supremum. Hence, in this case, sup A exists. From now on, suppose a = b, so a < b. We will inductively select a nested sequence of nonempty closed intervals I 1 ⊇ I 2 ⊇ ·· · ⊇ I n ⊇ · · · , whose intersection must then be nonempty by the Nested Interval Property, which we are assuming is true. We have already produced the first closed interval I 1 = [a, b]. We will explicitly describe the selection of the closed subinterval I 2 of I 1 and inductively explain how to select the closed subinterval I n+1 of I n following the same pattern. Consider the midpoint a+b m1 = of I 1 . If m1 is an upper bound for A, take I 2 = [a, m1 ]. If not, set 2 I 2 = [m1 , b]. In the first case, a ∈ I 2 implies that A ∩ I 2 = ∅ and the right hand endpoint of I 2 is an upper bound for A. In the second case, when m1 is not an upper bound for A, there is some element a ∈ A such that m1 < a . Yet b is a upper bound of A, so m1 < a ≤ b implies a ∈ I 2 , so again A ∩ I 2 = ∅ and the right hand endpoint of I 2 is an upper bound for A. In either case, I 2 = [a2 , b2 ] satisfies the following conditions: i. ii. iii. iv.
I 2 = [a2 , b2 ] is a nonempty closed subinterval of I 1 , the width of I 2 is b2 − a2 = 12 [b − a], b2 is an upper bound for A, and there is some element a ∈ A ∩ I 2 .
Now suppose we have the nested chain of closed intervals I 1 ⊇ I 2 ⊇ I 3 ⊇ · ·· ⊇ I n , where each I k = [ak , bk ] satisfies i. I k = [ak , bk ] is a nonempty closed subinterval of I k−1 , 1 b−a ii. the width of I k is bk − ak = [bk−1 − ak−1 ] = k−1 , 2 2 iii. bk is an upper bound for the set A, and iv. A ∩ I k =∅ an + bn for k = 2, 3, . . . , n. Consider I n and its midpoint m = , so an < m < bn . If m is 2 an upper bound for the set A, let I n+1 = [an , m]. Otherwise, take I n+1 = [m, bn ]. Then, writing I n+1 = [an+1 , bn+1 ], we still have an+1 < bn+1 , bn+1 is an upper bound for the set A, and A ∩ I n+1 = ∅ (by the same argument given in the construction of I 2 ). By induction, for each n ∈ N, there is a closed interval I n = [an , bn ] with the properties (i) through (iv) listed above. In particular, this is a nested sequence of nonempty closed intervals, so the Nested Interval Property implies there must be an element ∞
I . s∈ n
n=1
We claim that s is the supremum of the set A. To prove this, we will first show that s is an upper bound for A and then demonstrate that s must be the least upper bound using Lemma 1.3.7. Suppose that s were not an upper bound for A. Then, by definition, there must exist some element a ∈ A such that a > s. Take ε = a − s > 0 and consider the real b−a number . By the Archimedean Property, there is a natural number M ∈ N such ε b−a b−a that < M . Moreover, for all n ∈ N, n ≤ 2n−1 implies that < M ≤ 2M −1 , ε ε
Math 320, Real Analysis I
Solutions to Homework 3 Problems
b−a b−a ∈ − ≤ − < ε. Then s I implies that b s b a = < ε = a − s, so M M M M M −1 M −1 2 2 bM < a which contradicts that bM is an upper bound for A. Thus s is an upper bound for A. b−a Now, let ε > 0 be arbitrary. Consider the real number . By the Archimedean ε b−a b−a Property, there is a natural number K ∈ N such that < K ≤ 2K −1 , so K −1 < ε. ε 2 b−a Then s − aK ≤ bK − aK = K −1 < ε and I K = [aK , bK ] ∩ A = ∅ implies that there exists 2 a ∈ A such that a ≥ aK > s − ε, so s = sup A by Lemma 1.3.7. so
(b) Use the Monotone Convergence Theorem to give a proof of the Nested Interval Property. Proof. Let I 1 ⊇ I 2 ⊇ I 3 ⊇ · · · ⊇ I n ⊇ · · · be a nested sequence of nonempty closed
intervals, I n = [an , bn ] for all n ∈ N. Now consider the sequence (an ) of the left hand endpoints of the intervals I n . Since the intervals are nested, for each n we have [an+1 , bn+1 ] = I n+1 ⊆ I n = [an , bn ], so it follows that an+1 ≥ an and bn+1 ≤ bn for all n ∈ N. Therefore, our sequence (an ) is monotone increasing. Furthermore, (an ) is bounded, since for all n we have a1 ≤ an ≤ bn ≤ b1 . Therefore, by the Monotone Convergence Theorem , the sequence (an ) converges, so let α = lim an . We claim that α ∈ ∞ I , which will imply that this intersection is nonempty, and n=1 n thus prove the Nested Interval Property. Thus we must show that α ∈ I n = [an , bn ] for all n, which means we must demonstrate that an ≤ α ≤ bn is always true. First, because the intervals are nested, we see that every bm serves as an upper bound for the sequence (an ), so by the Order Limit Theorem we may conclude that α = lim an ≤ bm for every m. Moreover, α ≥ an for each n, for if α < aN for some N , then |an − α| = an − α ≥ aN − α > 0 for all n ≥ N since (an ) is increasing, which contradicts that α = lim an . Therefore, an ≤ α ≤ bn for every n, so α ∈ I n = [an , bn ] for each n ∈ N. Therefore, α∈ ∞ I , so this intersection is not empty. n=1 n
(c) Use the Bolzano-Weierstrass Theorem to prove the Nested Interval Property. Proof. Let I 1 ⊇ I 2 ⊇ I 3 ⊇ · · · ⊇ I n ⊇ · · · be a nested sequence of nonempty closed
intervals, I n = [an , bn ] for all n ∈ N. Now consider the sequence (an ) of the left hand endpoints of the intervals I n . Since the intervals are nested, for each n we have [an+1 , bn+1 ] = I n+1 ⊆ I n = [an , bn ], so it follows that an+1 ≥ an and bn+1 ≤ bn for all n ∈ N. Therefore, our sequence (an ) is monotone increasing. Furthermore, (an ) is bounded, since for all n we have a1 ≤ an ≤ bn ≤ b1 . Therefore, by the BolzanoWeierstrass Theorem , the sequence (an ) has a subsequence (an ) which converges, so let α = lim an . We claim that α ∈ ∞ I , which will imply that this intersection is nonempty, and n=1 n thus prove the Nested Interval Property. Thus we must show that α ∈ I n = [an , bn ] for all n, which means we must demonstrate that an ≤ α ≤ bn is always true. First, because the intervals are nested, we see that every bm serves as an upper bound for the sequence (an ), and hence also of the subsequence (an ), so by the Order Limit Theorem we conclude that α ≤ bm for every m. Moreover, as in Part (b), α ≥ an for each k, and we know ak ≤ an since k ≤ nk , so it follows that α ≥ ak for all k ∈ N. Therefore, an ≤ α ≤ bn for every n, so α ∈ I n = [an , bn ] for each n ∈ N. Thus, α ∈ ∞ I , so this n=1 n intersection is not empty. k
k
k
k
k
Math 320, Real Analysis I
Solutions to Homework 3 Problems
(d) Use the Cauchy Criterion to prove the Bolzano-Weierstrass Theorem. Proof. Let (an ) be a bounded sequence. Thus there exists a real number M > 0 such
that |an | ≤ M for all n ∈ N. That is, −M ≤ an ≤ M for every term an of the sequence, of which there are infinitely many. Thus, if we bisect the interval [ −M, M ], at least one of the subintervals [−M, 0] or [0, M ] must still contain infinitely many of the terms an of our sequence. Let I 1 be a half which contains infinitely many terms and select an1 to be one of the terms of our sequence that is in the interval I 1 . Notice that the width of the interval I 1 is M . Now divide I 1 in half and let I 2 be a half that contains infinitely many of the terms an . Select an2 ∈ I 2 such that n2 > n1 , which we can do since I 2 contains infinitely many an ’s and the requirement that n2 > n1 means we exclude only finitely many of these from consideration (which leaves us with infinitely many left) from which to select an2 . Then I 2 has width M/2. Continuing in this fashion, for each k, starting our current interval I k that possesses infinitely many terms an from the sequence (including an that we select in it), divide it in half and let I k+1 be a half that still contains infinitely many terms. Select an +1 ∈ I k+1 so that nk+1 > nk , which we may do since this requirement excludes at most nk of the infinitely many terms an in the subinterval I k+1 , which leaves infinitely many left from which to select an +1 . In this way, we have constructed a subsequence (an ) of (an ) and a nested sequence I 1 ⊇ I 2 ⊇ ·· · ⊇ I k ⊇ · · · of subintervals of [ −M, M ] such that the width of I k is given by M/2k−1 for each k. We claim that our subsequence (an ) is Cauchy, and hence convergent. Let ε > 0 be arbitrary. Consider the real number M/ε. By the Archimedean Property, there is a natural number K ∈ N such that M/ε < K , and we know that K ≤ 2K −1 , so it follows that M/2K −1 < ε. Let k, ≥ K and consider |an − an |. Now an ∈ I k ⊆ I K and an ∈ I ⊆ I K , so the greatest the distance between an and an can be is the width of I K , which is M/2K −1 , so M |an − an | ≤ K −1 < ε 2 whenever k, ≥ K . Therefore, by definition, our subsequence is a Cauchy sequence, so it converges by the Cauchy Criterion . Therefore, the bounded sequence (an ) has a convergent subsequence, which proves the Bolzano-Weierstrass Theorem. k
k
k
k
k
k
k
k
k
Exercise 2.7.6. (a) Show that if xn converges absolutely, and the sequence (yn ) is bounded, then the sum xn yn converges.
Proof. Let M > 0 be a real number such that |yn | ≤ M for all n ∈ N, which exists since
(yn ) is bounded. Let ε > 0 be arbitrary. Then ε/M > 0 as well, so there is a natural number N ∈ N such that
||xm+1 | + |xm+2 | + · · · + |xn || = |xm+1 | + |xm+2 | + · · · + |xn | <
ε M
|xn |, which whenever n > m ≥ N by the Cauchy Criterion for Series applied to converges since xn is absolutely convergent. Now consider the series xn yn . We will show that it is absolutely convergent, and thus converges by the Absolute Convergence Test. That is, we will show that the series
Math 320, Real Analysis I
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|x y | converges, so let n > m ≥ N and consider n n
||xm+1 ym+1 | + |xm+2 ym+2 | + · · · + |xn yn || = |xm+1 | |ym+1 | + |xm+2 | |ym+2 | + · · · + |xn | |yn | ≤ |xm+1 | M + |xm+2 | M + · · · + |xn | M = [|xm+1 | + |xm+2 | + · · · + |xn |] M ε < M = ε. M
|xn yn | converges by the Cauchy Criterion for Series , so Therefore, xn yn is absolutely convergent. Thus xn yn converges by the Absolute Convergence Test .
(b) Find a counterexample that demonstrates that Part (a) does not always hold if the convergence of xn is conditional. (−1)n+1 xn = , which is the Alternating Harmonic Series. We Example: Consider n already know that this series converges, but that its convergence is conditional. Now let (yn ) be the bounded sequence (−1)n+1 , which is bounded as |yn | = 1 ≤ 1 for all n. (−1)n+1 1 Then the series xn yn = (−1)n+1 = , which is the Harmonic n n Series, which we all know diverges. Hence the result of Part (a) need not hold if the convergence of xn is conditional.
Exercise 2.7.9. Ratio Test: Given a series
a
n
with an = 0, the Ratio Test states that if (an ) satisfies
a lim a
n+1 n
then the series converges absolutely.
= r < 1,
(a1 ) Why is there a number r satisfying r < r < 1? Solution: Since r < 1, the Density of Q in R implies that there is a rational number r such that r < r < 1. r+1 Another solution to this question is found by taking r = , the average of r and 1, 2 which must be strictly between them since r < 1. (a2 ) Show that there is a natural number N such that n ≥ N implies |an+1 | ≤ | an |r . Proof. Suppose the sequence (an ) of terms of the series
a lim a
n+1 n
= r < 1.
a
n
satisfies
Let r be a real number such that r < r < 1. Then set ε = r − r, which is positive since an+1 r < r . Since lim exists, for our ε > 0, there exists a natural number N ∈ N such an that an+1 − r < ε = r − r an
for all n ≥ N . Hence, whenever n ≥ N ,
a a
n+1 n
∈ (r − [r − r], r + [r − r]) = (2r − r , r ),
Math 320, Real Analysis I
a so a
n+1 n
Solutions to Homework 3 Problems
a < r when n ≥ N . Now a
n+1
all n. Therefore, when n ≥ N , we have
n
|a | = |a | and a n+1
n
= 0 implies |an | > 0 for
n
|an+1 | < r , so |an+1 | < |an |r as we needed to |an |
show. (b) Why does |aN |
n
(r )
necessarily converge?
a By the Order Limit Theorem, 0 ≤ r = lim , so 0 ≤ r < r < 1, which a (r ) is a geometric series with ratio r and |r | < 1, so | | implies that r = r < 1. Then (r ) converges by Example 2.7.5. Finally, by the Algebraic Limit Theorem for Series, the series |a | (r ) = |a |(r ) converges as well. (c) Show that |a | converges. a = r < 1. By Let a be a series with a = 0 for all n such that lim a Solution:
n+1
n
n
n
N
n
N
n
n
Proof.
n
n+1
n
n
Part (a), there is a real number r with r < r < 1 and natural number N ∈ N such that |an+1 | ≤ |an |r whenever n ≥ N . We claim that for all k ∈ N, |aN +k | ≤ |aN |(r )k . When k = 1, this follows from Part (a) as mentioned above. Now suppose it is true that |aN +k | ≤ | aN |(r )k and consider
|aN +(k+1) | ≤ |aN +k
|r ≤ |a
|(r ) r = |a
N
k
N
|(r )k+1 .
Thus, by induction, |aN +k | ≤ | aN |(r )k for all k ∈ N. |an |, which we need to show is convergent. First, we break Now consider the series the infinite series into two parts, a finite initial portion and an infinite tail, as
N
∞
n
n=1
N
∞
|a | = |a | + n
n=1
n
n=N +1
∞
|a | + |a |a | =
N +k
n
n=1
|.
k=1
|an | is finite, so ∞ |an | converges if and only if ∞ |aN +k | Clearly the finite sum N n=1 n=1 k=1 converges. To prove that k |aN +k | converges, we employ the Comparison Test , as 0 ≤ | aN +k | ≤ | aN |(r )k for all k ∈ N and k |aN |(r )k converges by Part (b) implies that |aN +k |, converges as well. Therefore, the series ∞ |an | converges. our series, ∞ k=1 n=1
