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Math 323 – Real Anlaysis – Homework 4 Solutions
2.5.3a Take xn = (^1) n if n is even and xn = n+1 n if n is odd.
2.5.3b This is impossible since by the Monotone Convergence Theorem a monotone divergent sequence must be unbounded. And thus every subsequence must also be unbounded.
2.5.3c Choose the elements of a sequence, (xn), from the following list, starting at the top left and following the arrows: ↙ ↙ ↙ ↙ 1 1 1 1 ...
1 2
1 2
1 2
1 2 ... 1 3
1 3
1 3
1 3 ... 1 4
1 4
1 4
1 4 ... 1 5
1 5
1 5
1 5 ... .. .
So the first few elements of the sequence will be (xn) = (1, 1 , 12 , 1 , 12 , 13 , 1 , 12 , 13 , 14 , ...)
2.5.3d (an) = (1, 2 , 1 , 3 , 1 , 4 , 1 , 5 , 1 , 6 , ...)
2.5.3e This is impossible since the Bolzano-Weierstrass Theorem implies that any bounded sequence will have a convergent subsequence.
2.5.6 Let (an) be a bounded sequence and define the set
S = {x ∈ R : x < an for infinitely many terms an}
Note that the set S is bounded above, since (an) is bounded and any upper bound of the sequence will be an upper bound of S. Therefore, sup S exists, define s = sup S. Now, consider an arbitrary ε > 0. Since s + ε cannot be an element of the set S, there must be only a finite number of an such that an > s + ε. But by the definition of the supremum, for any ε > 0 there exists some x ∈ S with s − ε < x. Thus, for any ε > 0 there are an infinite number of terms an with s − ε < an and only a finite number of those terms satisfy an > s + ε. Therefore, we must have an infinite number of an with s − ε < an < s + ε. Now, to construct a subsequence (ank ) → s, consider εk = 1/k. So, start by choosing some an 1 so that s − 1 < an 1 < s + 1, from this point on, we choose ank+1 so that nk+1 > nk and s − 1 /k < ank+1 < s + 1/k. We know that we can always do this, since at every step k there are an infinite number of an with s − 1 /k < an < s + 1/k.
2.6.1a The sequence (an) with an = (−1)n 1 /n is not monotone, but converges to 0 and thus must be a Cauchy sequence.
2.6.1b The sequence (n) is monotone increasing but the distance between the terms is always greater than or equal to 1.
2.6.1c This is not possible since every Cauchy sequence converges by the Cauchy Criterion and by Theorem 2.5.2 every subsequence must also converge to the same limit.
2.6.1d Let (an) be defined as follows
an =
n n is even 1 /n n is odd
2.6.3a The pseudo-Cauchy property only specifies that you can control the distance between consecutive terms, while the Cauchy property specifies the distance between all terms past a certain N.
2.6.3b Consider the sequence of partial sums (sm) where sm =
n=1 1 /n.^ The difference between consecutive terms is 1/n, so that the sequence satisfies the pseudo-Cauchy property, but this is the harmonic series which we have shown is divergent and thus cannot be Cauchy.
2.6.6a Assume that the Nested Interval Property holds and let A be any set that is bounded above and non-empty. Then if M is any upper bound of A and x ∈ A we will search for the supremum of A in the interval [x, M ]. We begin by defining a sequence of nested intervals, In, as follows: Let I 0 = [x, M ]. Then split I 0 into two halves. If the right half contains an element of A then define the interval I 1 to be the right half, otherwise define I 1 to be the left half of the interval. Proceed inductively, defining Ik to be the right half of the interval Ik− 1 if there is an element of A in that interval, otherwise let Ik be the left half. Note that at each step the interval Ik contains some element of A. Further, the lengths of the intervals are converging to zero: |Ik| = M 2 −n x. By the Nested Interval Property the intersection ∩∞ n=1Ik 6 = so let s ∈ ∩∞ n=1Ik. We will show that s = sup A. By our construction of the intervals Ik there is no element of A that lies above any interval, since we always choose the interval to the right if it contains elements of A. Thus, s is an upper bound of the set A. Further, for any ε > 0 there exists some N such that |IN | < ε and since IN contains some element of a ∈ A and s ∈ IN we have that |a − s| < |IN | < ε. Thus, we have satisfied the definition of supremum and shown that s = sup A.
2.7.4 Take both xn = yn = 1/n then individually
xn and
∑ yn^ diverge, but the product xnyn =
1 /n^2 converges.
