Math 163 (23) - Midterm Test 1

Size: px

Start display at page:

Download "Math 163 (23) - Midterm Test 1"

Joel Harvey
5 years ago
Views:

Name: Id #: Math 63 (23) - Midterm Test Spring Quarter 208 Friday April 20, 09:30am - 0:20am Instructions: Prob. Points Score possible 26 2 4 3 0 TOTAL 50 Read each problem carefully. Write legibly.

1 Name: Id #: Math 63 (23) - Midterm Test Spring Quarter 208 Friday April 20, 09:30am - 0:20am Instructions: Prob. Points Score possible TOTAL 50 Read each problem carefully. Write legibly. Show all your work on these sheets. Feel free to use the opposite side. This exam has 5 pages, and 3 problems. Please make sure that all pages are included. You may not use books, notes, calculators, etc. Cite theorems from class or from the texts as appropriate. Proofs should be presented clearly (in the style used in lectures) and explained using complete English sentences. Good luck!

2 Math 63 (23) - Midterm Test Spring Quarter 208 Page 2 of 5 Question. (Total of 26 points) a) (6 points) State the definition of a limit of a sequence. Solution. A sequence (a n ) converges to a limit l if for all ε > 0 there exists some N N such that a n l < ε for all n > N. b) Consider the recursively defined sequence (a n ) given by a := ; a n+ = a n + 2 for all n N. 3 i) (3 points) Show that the sequence is bounded above by 3. ii) (4 points) Show that the limit l of the sequence exists. iii) (3 points) Using the previous part and the recursive formula, conclude that l = 3. Solution. i) We prove the claim using induction on n. For n = we clearly have a = < 3. Supposing a n 3, it follows that a n+ = a n / /3 + 2 = 3, as required. Hence, by the principle of mathematical induction, a n 3 for all n N. ii) By part i) and the monotone convergence theorem for sequences, it suffices to show the sequences is non-decreasing. Let n N and note that a n+ a n = 2 2a n / /3 = 0, since a n 3. iii) Since lim a n = l exists, taking the limit in the identity a n+ = a n /3 + 2 yields l = lim a n+ = lim Thus, l = 3, as required. a n = l c) Determine whether the following series converge i) (5 points) ii) (5 points) n=2 n log n ; 0 n n!.

3 Math 63 (23) - Midterm Test Spring Quarter 208 Page 3 of 5 Solution. i) We give two proofs: ) We invoke the integral test. By a simple substitution (u = log x), it follows that for R e one has R e dx log R x log x = du u = log log R. Since log : (0, ) R is unbounded, it follows that log log: (, ) is also unbounded and so the improper integral e dx x log x diverges. By the integral test, the series also diverges. 2) By the Cauchy condensation test (as discussed in the problem session), the series converges if and only if k= 2 k 2 k log 2 k = log 2 converges. But since the right-hand side is a scalar multiple of the harmonic series, we see it must diverge. ii) We use the ratio test. Let a n := 0 n /n! so that a n+ a n = 0n+ n! 0 n (n + )! = 0 n + Hence, by the ratio test the series converges. k= k. 0 as n.

4 Math 63 (23) - Midterm Test Spring Quarter 208 Page 4 of 5 Question 2. (Total of 4 points) i) (4 points) Let a < b and a n [a, b] for all n N. Suppose lim a n = l. Show l [a, b]. ii) (4 points) State the Bolzano Weierstrass theorem. iii) (6 points) Using the Bolzano Weierstrass theorem, prove the familiar result that if f : [a, b] R is continuous, then it is bounded above. (Hint: suppose f is not bounded above and use this assumption to construct a relevant sequence (a n ) with a n [a, b] for all n N.) Solution. i) Suppose l > b and let ε := (l b)/2. Then there exists some N N such that a n l < ε for all n > N. But a n l = l a n l b = 2ε, a contradiction. Hence l b. By a similar argument, l a. ii) Every bounded sequence admits a convergent subsequence. iii) Following the hint, suppose f : [a, b] R is not bounded above so that for every n N one can find some a n [a, b] with f(a n ) n. By the Bolzano- Weierstrass theorem there exists a convergent subsequence (a nk ) k= of the (a n) with lim a nk = l, say. By part i) we know that l [a, b]. Note that f(a nk ) n k k so the sequence (f(a nk )) k= is unbounded and therefore divergent. Hence, f cannot be continuous at l.

5 Math 63 (23) - Midterm Test Spring Quarter 208 Page 5 of 5 Question 3. (Total of 0 points) a) (6 points) State what it means for the series i) Converge. ii) Converge absolutely. iii) Converge conditionally. a n to: N Solution. i) The sequence of partial sums (s N ) N= given by s N := a n converges. ii) The series a n converges. iii) a n converges but does not converge absolutely. b) (4 points) Determine whether the series ( ) n + log 2 n converges. Solution. Since /( + log 2 n) is non-negative and decreasing for n and lim ( + log 2 n) = 0, it follows by the alternating series test that the series converges.

Math 256 (11) - Midterm Test 2

Math 256 (11) - Midterm Test 2 Name: Id #: Math 56 (11) - Midterm Test Spring Quarter 016 Friday May 13, 016-08:30 am to 09:0 am Instructions: Prob. Points Score possible 1 10 3 18 TOTAL 50 Read each problem carefully. Write legibly.