Download Practice Final Exam Solutions - Analysis I | MATH 414 and more Exams Mathematics in PDF only on Docsity! Math 414 Professor Lieberman May 6, 2003 SOLUTIONS TO PRACTICE FINAL EXAM 1. Define a new sequence (bn) by bn = an+1− an. Then limn→∞(bn + bn−1) = 0. Given ε > 0, there is a natural number N such that, if n ≥ N , then |bn + bn−1| < ε/2. We now prove by induction (on k) that |bN+k| ≤ k ε 2 + |bN |. For k = 1, we have |bN+1| = |bn − (bN + bN+1)| ≤ |bN |+ |bN + bN+1| ≤ ε 2 + |bN |. If the inequality is true for k = m, then |bN+(m+1)| = |bN+m − (bN+m + bN+(m+1)| ≤ |bN+m|+ |bN+m + bN+(m+1)| ≤ mε 2 + |bN |+ ε 2 ≤ (m + 1)ε 2 + |bN |. Now take M to be a natural number greater than N such that |bN |/M < ε/2. If n ≥ M , then ∣∣∣∣bnn ∣∣∣∣ ≤ n−Nn ε2 + |bN |n < ε. It follows that lim n→∞ bn/n = 0. 2. For n = 1, we have 1∑ k=1 Fk = F1 = 1 = F1+2 − 1. If the statement is true for n = m, then m+1∑ k=1 Fk = m∑ k=1 Fk + Fm+1 = Fm+2 − 1 + Fm+1 = Fm+3 − 1. 3. In both parts, we will use the interval [0, 1]. (a) Use f(x) = 1 2 if x = 0, x if 0 < x < 1, 1 2 if x = 1. (b) Use f(x) = 1 if x = 0, x if 0 < x < 1, 0 if x = 1. 1 2 4. By algebra, a2f(x + bh)− b2f(x + ah) + (b2 − a2)f(x) (a2b− b2a)h = a2[f(x + bh)− f(x)] (a2b− b2a)h − b 2[f(x + ah)− f(x)] (a2b− b2a)h , so lim h→0 a2f(x + bh)− b2f(x + ah) + (b2 − a2)f(x) (a2b− b2a)h = a2 a2 − ba lim h→0 f(x + bh)− f(x) bh − b 2 ab− b2 lim h→0 f(x + ah)− f(x) ah = ( a2 a2 − ba − b 2 ab− b2 ) f ′(x) = f ′(x). 5. The substitution u = ex (so x = ln u and dx = du/u) gives∫ ∞ 0 sin(ex) dx = ∫ ∞ 1 sinu u du, which converges. (More formally, we would say that∫ ∞ 0 sin(ex) dx = lim b→∞ ∫ b 0 sin(ex) dx = lim b→∞ ∫ ln b 1 sinu u du, which exists because the improper integral∫ ∞ 1 sinu u du converges.) 6. Because the sequence (bk) is bounded, there is a number M such that |bk| ≤ M for all k. Because ∑ |ak| converges, given ε > 0, there is a natural number N such that, if n ≥ m ≥ N , then n∑ k=m |ak| ≤ ε M . Now, if n ≥ m ≥ N , then n∑ k=m |bkak| ≤ n∑ k=m M |ak| = M n∑ k=m |ak| < ε, which means that ∑ akbk converges absolutely. 7. This is an alternating series, so set ak = 1/(k ln k). Then ak > 0, (ak) is a decreasing sequence, and ak → 0 as k → ∞. Therefore the series converges by the alternating series test. 8. FALSE. Take f(x) = { 1 if x = 0, 1 x if x > 0, and define xn = 1/n. Then lim n→∞ f(xn) = ∞, but lim x→∞ f(x) = 0.
