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Daepp and Gorkin, Solutions to Reading, Wrting, and Proving, Chapter 20 Reading, Writing, and Proving (Second Edition) A Note to Student Users. Check with your instructor before using these solutions. If you are expectedto work without any help, do not use them. If your instructor allows you to find help here, then we giveyou permission to use our solutions provided you credit us properly.
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Solution to Problem 20.3. Proof. From Problem 5.14 we conclude that |x − a| < δ if and only if−δ < x − a < δ. The latter inequalities are equivalent to a − δ < x < a + δ. This proves the statement.
(a) In Problem 18.5 (b) we found that S1 = 75 and Sn+1 = 0.65Sn + 75.
(0.65)75(1 + 0.65) + 75 = 75(1 + 0.65 + 0.652) Using the formula for a geometric sum we guess that Sn = 75(1 − 0.65n)/0.35 for n ≥ 1.
Proof. We establish the claim with an inductive proof.
The case n = 1: S1 = 75(1 − 0.65)/0.35 = 75, which is correct.
Let n ≥ 1 and suppose that Sn = 75(1 − 0.65n)/0.35. Then Daepp and Gorkin, Solutions to Reading, Wrting, and Proving, Chapter 20 (b) We claim that limn→∞ Sn = 75/0.35.
Proof. We first show that (0.65)n → 0.
To see this, let ε > 0. We may also assume ε < 1. Let N = (ln ε)/(ln 0.65). Note that N > 0. Forn > N we have Using Theorem 20.9, we now see that Sn → 75/0.35.
(c) limn→∞ sn = limn→∞(Sn − 75) = limn→∞ Sn − 75 = 75 − 75 = 975 .
(d) After one week the amount of phenytoin in the patient’s blood is between s14mg and S14mg. This can be shown to be between 138.77mg and 213.77mg.
After a full month, the level is between s60 and S60; that is, between 139.29mg and 214.29mg.
(e) In the long run the patient’s phenytoin level is between limn→∞ sn and limn→∞ Sn, which is between Note that there is no big difference, the phenytoin level is already stable after one week.
(c) We note that the sequence (1/ n) is decreasing and bounded below. By Theorem 20.12 this sequence converges. So limn→∞ (1/ n) = a for some real number a.
By Theorem 20.9 (iii), we have (limn→∞ (1/ n)) = a2 = limn→∞ (1/n). From Example 20.2 weconclude that a2 = 0. Hence a = 0.
n + 7 < 1/ n. Problem 20.8 (a) implies that limn→∞ √ Daepp and Gorkin, Solutions to Reading, Wrting, and Proving, Chapter 20 (e) Note that using Theorem 20.9 and Example 20.2 as above, we can show that limn→∞ 2 + 1 = 2 and (g) Note that 0 < n < n + 7 < (n + 7)! for all integers n ≥ 1. Hence 0 < = 0. It follows from Problem 20.8 (a) that lim (h) For this last part we leave it to you to fill in the details of the use of Theorem 20.9 and Example 20.2: (a) From Problem 5.14 we know that for all real numbers xn we have −|xn| ≤ xn ≤ |xn|. Adding |xn| gives the required inequality (b) As pointed out in part (a), we have −|xn| ≤ xn ≤ |xn| for all n. The result of Problem 20.8 (b) Daepp and Gorkin, Solutions to Reading, Wrting, and Proving, Chapter 20 (c) The answer to this question is no. Consider xn = (−1)n for n ∈ N. Then |xn| = 1 and thus |xn| → 1.
In Exercise 20.6 you showed that (xn) does not converge.
(a) Since 0 < a < 1, we have 1 − a > 0 and x = (1 − a)/a is a positive (b) Note that since x > 0, we have 1 + x > 0. Bernoulli’s inequality (Problem 18.6) applies and hence (1 + x)n ≥ 1 + nx for all n ∈ N. Hence an = 1/(1 + x)n ≤ 1/(1 + nx) for all n ∈ N.
(c) Recall that x > 0. Hence for all n ≥ 1 we have that Using Example 20.2, Theorem 20.9 (ii), and Problem 20.8 (a) we conclude that 1/(1 + nx) → 0.
(d) From Part (b) we have 0 < an ≤ 1/(1 + nx) for all n ∈ N. Part (c) and Problem 20.8 (a) imply that (e) Note that we can write xn alternatively as xn = 1 − (1/10)n. It now follows from Part (d) above and the rules of Theorem 20.9, that xn → 1.
Note that this justifies the equality 0.99999 . . . = 1 and shows that the decimal representation of thereal number 1 is not unique.
n+1 = xn + 9 · 10−(n+1) > xn and xn < 1 for all n ∈ Z . Hence (xn) is an increasing and bounded sequence. By Theorem 19.10 it converges to its supremum. In Problem19.14 we showed that limn→∞ xn = 1. Hence sup(xn) = 1.
Solution to Problem 20.21. We leave this one to you.