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Solutions to Practice Problems, Math 220 1 Evaluate the following limits: 2n 1 lim 1 + and n→∞ n Solution n 1 lim 1 − . n→∞ n 2n n 2 1 1 1+ = 1+ n n and so 2n 1 lim 1 + = e2 . n→∞ n Similarly, n n −1 1 n 1− = n n−1 and so lim n→∞ 1 1− n n = e−1 . 2 Use the definition of limit to evaluate n a: limn→∞ 2 + n1 b: limn→∞ 3nn c: limn→∞ sin(n)+cos(n) 3 n Solution n a: We claim that the sequence diverges. If limn→∞ 2 + n1 = x for some real number x, then since n n √ 1 1 n =2 1+ < 2n · e 2+ n 2n and 2+ 1 n+1 n+1 > 2n+1 = 2n · 2, we have n+1 n √ 1 1 2+ − 2+ > (2 − e) · 2n > 0.7. n+1 n If we take = 0.35, then there exists N such that n > N implies n 1 2+ − x < 0.35. n 1 2 On the other hand, n+1 n n n 1 1 1 1 2+ ≤ 2+ − 2+ −x + x− 2+ n+1 n n n and this is less than 0.35 + 0.35 = 0.7, a contradiction. b: We have limn→∞ 3nn = 0. To see this, note that 3nn < 2−n for all n ∈ N (you can show this by induction). If we thus take N = − log()/ log(2), then n > N implies that n n − 0 = n < 2−n < 2−N = , n 3 3 as desired. n sin(n)+cos(n) c: limn→∞ = 0. Here, we use the inequalities 3 −2 < sin(n) + cos(n) < 2, which lead to −(2/3) < n sin(n) + cos(n) 3 n < (2/3)n . Taking N = − log()/ log(3/2), and assuming n > N implies that n sin(n) + cos(n) − 0 (2/3)n < (2/3)N = . 3 3 For each of the following sequences, determine whether or not the sequence converges or diverges. Prove your answer. a: a1 = 1, an+1 = an + a1n √ b: b1 = 3, bn+1 = 1 + bn − 1 Solution a: If lim an = a, then we would have a = a + a1 and so 0 = 1/a, a contradiction. It follows that an diverges. b: We claim that bn ≤ 3. To see this, use induction. Since b1 = 3, let use suppose that bk ≤ 3. Then p √ bk+1 = 1 + bk − 1 ≤ 1 + 2 < 3. Similarly, we have bn > 2. To see this, again use induction. Next we claim that bn is decreasing. Indeed, if bn+1 /bn > 1 √ √ then 1 + bn − 1 > bn or bn − 1 < bn − 1, contradicting bn > 2. It follows, via the Monotone Convergence Theorem √ that bn converges, say to b. Since we have 1 + b − 1 = b and b ≥ 2, we conclude that lim bn = 2. P P 4 Suppose that an converges. Either prove that bn converges or give a counterexample, where a: bn = an /n 3 √ b: bn = an /n, an ≥ 0 c: bn = an sin(n) d: bn = n1/n an Solution Upon reflection, you can skip this question, which involves rather more of the chapter on series than we’ve covered. √ 5 If xn = 3 n, show that (xn ) satisfies lim |xn+1 − xn | = 0, but that it is not a Cauchy sequence. Solution Since √ √ 3 n+1− 3n = 1 1 < ( ( ( (n + + (n(n + 1)) 1/3) + n 2/3) 3n 2/3) √ √ √ 3 3 it follows that lim n + 1 − n = 0. On the other hand, since 3 n is unbounded, it fails to converge and hence is not a Cauchy sequence. 1)( 2/3) 6 Prove that the set S of odd integers is countable. Solution As a subset of Z, S is countable.