Homework 4, 5, 6 Solutions. > 0, and so a n 0 = n + 1 n = ( n+1 n)( n+1+ n) 1 if n is odd 1/n if n is even diverges.
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1 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 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 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 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 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 = (c) lim = 0 by Theorem n 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 =.
3 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 = (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 (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 (a) Is it possible to have an unbounded sequence {a n } such that lim a n/n = 0? Yes. Example a n = n (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 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 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 (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 Give an example of a sequence that diverges to + but is not eventually increasing. a n = n + ( ) n 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 = = 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 = If we know an 30, then a n+ = 6 + a n = 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 = (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 = 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 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)
5 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 (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 (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 (c) Give an example of a set S where both sup S and exactly one accumulation point exist, but the values are not equal (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
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