2..2(a) lim a n = 0. Homework 4, 5, 6 Solutions Proof. Let ɛ > 0. Then for n n = 2+ 2ɛ we have 2n 3 4+ ɛ 3 > ɛ > 0, so 0 < 2n 3 < ɛ, and thus a n 0 = 2n 3 < ɛ. 2..2(g) lim ( n + n) = 0. Proof. Let ɛ > 0. Then for n n = we have 2 n 4ɛ 2 ɛ > 0, and so a n 0 = n + n = ( n+ n)( n++ n) n++ n = n++ n < 2 n ɛ. 2..2(k) The sequence a n = { if n is odd /n if n is even diverges. Proof. Assume not. Then the sequence converges to some limit A R. By definition of convergence (with ɛ = /4) there exists n such that a n A < /4 for n n. Choose an integer k n /2. Then 2k n and 2k + n, so a 2k A < /4 and a 2k+ A < /4. So /(2k) A < /4 and A < /4. The second inequality implies A > 3/4, and the first one A < /(2k) + /4 /2 + /4 = 3/4. This is a contradiction, so the statement is proved. 2..7 If lim a 2n = A and lim a 2n = A, then lim a n = A. Proof. Let ɛ > 0. Then there exist n and n 2 such that a 2n A < ɛ for n n and a 2n A < ɛ for n n 2. Let n = max(2n, 2n 2 ), and let n n be arbitrary. If n is even, then there exists k N such that n = 2k. Since n n 2n we get that k n, and thus a n A = a 2k A < ɛ. If n is odd, then there exists k N such that n = 2k. Since n n 2n 2, we get k n 2, and thus a n A = a 2k A < ɛ. Every number n is either even or odd, so we have proved the claim. The converse is also true: If lim a n = A, then lim a 2n = lim a 2n = A. Proof. Let ɛ > 0. Then there exists n such that a n A < ɛ for n n. For n n = (n + )/2 we get 2n 2n n and thus a 2n A < ɛ and a 2n A < ɛ. This shows the claim. The first direction helps with 2(j), since lim a 2n = lim /(2n) = 0 and lim a 2n = 0, so the results implies that the sequence converges to 0.
2 The converse helps with 2(k), since lim a 2n = lim /(2n) = 0 and lim a 2n =. If the sequence would converge, these two limits would have to be the same. Since they are different, the sequence itself diverges. 2..0 If {a n } converges to A, then the sequence {b n } defined by b n = (a n + a n+ )/2 converges to A, too. Proof. Let ɛ > 0. Then there exists n such that a n A < ɛ for n n. Then b n A = (a n + a n+ )/2 A = (a n A) + (a n+ A) /2 a n A /2 + a n+ A /2 < ɛ/2 + ɛ/2 = ɛ for n n. 2..20 Consider sequences {a n } and {b n }, where b n = n a n. (a) If {b n } converges to, does the sequence {a n } necessarily converge? No. Example is a n = n, b n = n n. (b) If {b n } converges to, does the sequence {a n } necessarily diverge? No. Example is a n = b n =. (c) Does {b n } have to converge to? No. Example is a n = b n = 0. 2.2.(c) lim = 0 by Theorem 2..3. 2n r n 2.2.(d) lim n! = 0. Proof. We will write a n = rn n!. Choose n N with n r. Let M = a n = r n = r n! r 2 r. n We first claim that a n M for all n n. Proof by induction: The case n = n is immediate by definition of M. Now if we already know the claim for some n n, then a n+ = r n+ (n + )! = r n n! r n + M r n + M. (We used n+ n r in the last inequality, and the induction hypothesis in the second-to-last inequality.) Now let n > n be arbitrary. Then n n, so r n (n )! M, and thus a n = r n n! = r n (n )! r n M r n. M r We know that lim n = 0. The squeeze theorem then implies that {a n } converges to 0, too. n 2.2.(i) lim n + n =.
Proof. This follows from the squeeze theorem, the estimate n n + n n 2n = n 2 n n n, and lim 2 = lim n n =. 2.2.3(a) Suppose that {a n } and {a n b n } both converge, and a n 0 for large n. Is it true that {b n } must converge? No. Example: a n = /n, b n = n. 2.2.3(b) Suppose that {a n } converges to a non-zero number and {a n b n } converges. Prove that {b n } must also converge. Proof. This follows immediately from the limit theorems. Let A = lim a n and C = lim a nb n. Then b n = anbn a n is the quotient of two convergent sequences, where the denominator converges to a non-zero limit. From Theorem 2.2.(c) we get that {b n } converges to C/A. 2.2.8(a) Is it possible to have an unbounded sequence {a n } such that lim a n/n = 0? Yes. Example a n = n. 2.2.8(b) Prove that if the sequence {a n } satisfies lim a n/n = L 0, then {a n } is unbounded. Proof. Assume not. Then there exists M such that a n M for all n, and thus a n /n M/n. We know lim M/n = 0, and the squeeze theorem implies lim a n/n = 0, contradicting the assumption. 2.2.2 If 0 α β, then lim n α n + β n = β. Proof. This is again the squeeze theorem. We know β n α n + β n 2β n, so β n α n + β n n n 2 β. In class we proved lim 2 =, so the sequence in the middle of the inequalities also has to converge to β. 3 2.3. Prove the Comparison Theorem: If {a n } diverges to +, and a n b n for n n, then {b n } also diverges to +. Proof. Let M > 0 be arbitrary. Then there exists n 2 such that a n > M for n n 2. For n n = max(n, n 2 ) we get b n a n > M. 2.3.3(a) Prove that a n = (n 2 + )/(n 2) diverges to +. Proof. For n 3 we have a n n 2 /n = n, and we already know that lim n = +. By comparison theorem {a n} diverges to +, too.
4 2.4. Give an example of a sequence that diverges to + but is not eventually increasing. a n = n + ( ) n. 2.4.2 Give an example of a converging sequence that does not attain a maximum value. a n = /n. 2.4.(a) Let the sequence {a n } be recursively defined by a = 6 and a n+ = 6 + a n for n N. Find the limit if it exists. First claim: The sequence is increasing. Proof by induction: a 2 = 6 + 6 6 = a. Assume that we know a n+ a n. Then a n+2 = 6 + a n+ 6 + an = a n+. Second claim: The sequence is bounded by 30. Proof by induction: a = 6 30. If we know an 30, then a n+ = 6 + a n 6 + 30 = 6 30. We know that monotone bounded sequences converge, so there exists some limit A R. We can pass to the limit in the recursive equation to get A = lim a n+ = lim 6 + an = 6 + A by limit theorems. From the equation A = 6 + A we see that A 0, and that A 2 = 6 + A. The two solutions of this equation are 2 and 3, and since A 0, we get A = 3. 2.4.(g) Same question as previous problem for a = and a n+ = + a n /2. First claim: The sequence is increasing. Proof by induction: a 2 = + /2 = a. Assuming that we know a n+ a n, we get a n+2 = + a n+ /2 + a n /2 = a n+. Second claim: The sequence is bounded by 30. Proof by induction: a = 30. Assuming we know a n 30, we get a n+ = + a n /2 + 30/2 = 6 30. Again we know that the sequence converges to some limit A R because it is monotone and bounded. Passing to the limit in the recursive equation we get A = lim a n+ = lim ( + a n/2) = + A/2 by limit theorems. The equation A = + A/2 has only one solution A = 2, so the limit is 2. 2.5. Let s 0 be an accumulation point of S. Prove that the following two statements are equivalent. (a) Any neighborhood of s 0 contains at least one point of S different from s 0. (b) Any neighborhood of s 0 contains infinitely many points of S. The statement of this problem is unfortunately slightly screwed up. Kosmala assumes from the outset that s 0 is an accumulation point. Then (a) is always true, since it is the definition of accumulation points. So the direction (b)
implies (a) is trivial, since (a) is true. It would also make both of these statements equivalent to completely unrelated true statements such as (c) The angle sum in a Euclidean triangle is π. We will show that (a) is equivalent to (b) for any s 0 R, without any assumptions about s 0 being an accumulation point. Proof. (b) = (a): Let ɛ > 0 be arbitrary. Then (s 0 ɛ, s 0 + ɛ) contains infinitely many points of S, so it contains at least two different points s, s 2 S. If s s 0, we have found a point of S in the neighborhood different from s 0. Otherwise we have s 2 s = s 0, so s 2 is the desired point. (a) = (b): Assume not. Then there exists ɛ > 0 such that the neighborhood (s 0 ɛ, s 0 + ɛ) contains only finitely many points of S. Denote this finite set of points by T, and let D = { s s 0 : s T, s s 0 } {ɛ}. The set D is finite (since T is finite) and non-empty (ɛ D), and all the numbers in D are positive. So ɛ = min D > 0 exists. Claim: There are no points of S different from s 0 in the neighborhood (s 0 ɛ, s 0 + ɛ ). Assume not. Then there exists s S (s 0 ɛ, s 0 + ɛ ) with s s 0. This implies 0 < s s 0 < ɛ ɛ, so s T. By definition s s 0 D, and thus ɛ s s 0. However, this contradicts s s 0 < ɛ. This contradiction proves the claim, and the claim itself contradicts (a), finishing the proof of this direction. 2.5.3 (a) Give an example of a sequence for which the set S = {a n : n N} has exactly two accumulation points. a n = ( ) n ( /n), accumulation points and. 2.5.3 (b) Give an example of a set S that contains infinitely many points but not every point of S is an accumulation point of S. 2.5.3 (c) Give an example of a set S where both sup S and exactly one accumulation point exist, but the values are not equal. 2.5.3 (d) Give an example of a set S where inf S and sup S are in S, but the accumulation point (points) is (are) not. Example for (b), (c), and (d): S = {( ) n /n : n N}. This set contains infinitely many points, the only accumulation point 0 is not in S, and both sup S = /2 and inf S = are in S. 5