G12MAN Mathematical Analysis Fully-justified answers to Revision Quiz questions Remember that, unless otherwise specified or not applicable for some reason), you MUST justify your answers to questions in this module with an argument sufficient to convince Dr Feinstein. Such arguments may sometimes be fairly concise, as long as they include the key points. In a typical G12MAN examination question, the justification is worth MORE than the rest of the answer. In the below, optional additional or alternative justification or comments often appear inside square brackets, [... ]. 1) Answer A sequence of real numbers is an infinite list of real numbers. A series of real numbers is an infinite sum, where each of the terms in the sum is a real number. Justification Not applicable. 2) a) Answer: x n 1/n converges to 0. Justification: In this case, I would accept the statement This is a standard convergent sequence from G11ACF. As a further revision exercise, you may wish to write down a full proof using the definition of convergence in terms of ε and N.) b) Answer: x n n2 + 2n + 3 converges to 1/2. 2n 2 + n + 1 Justification: n 2 + 2n + 3 2n 2 + n + 1 1 + 2 + 3 n n 2 2 + 1 + 1 n n 2 1 + 0 + 0 2 + 0 + 0 1 2 by the algebra of its, as n. c) Answer: x n n + 1) 2 n 2 diverges. Justification: n + 1) 2 n 2 2n + 1, and it is clear that this diverges to + as n. [Or, optionally: Since x n 2n + 1 > n, it is clear that x n diverges to + as n. ] d) Answer: x n + 1 converges to 0. Justification: For n N, we have n + 1 + 1 ) ) + 1) 2 ) 2 n + 1) n 1 1
and it is clear that this tends to 0 as n. e) Answer: x n 1 + 1 n )n converges to e. Justification: In fact, for the purposes of G12MAN you can quote that this is a standard convergent sequence from G11ACF. However, as a revision exercise from G11ACF, you may wish to be reminded of the proof. We take logs and look at ln x n n ln1 + 1 n ). [We would like to investigate the it of ln x n using L Hôpital s rule. Unfortunately, that rule does not apply to sequences, only to function its.] To help our investigation, we consider the following function it: x ln 1 + 1 ). x x Substituting y 1/x, we consider ln1 + y). y 0+ y This is an indefinite form of type 0/0. L Hôpital s rule tells us to differentiate numerator and denominator with respect to y and consider 1/1 + y). y 0+ 1 This it is clearly equal to 1. Thus, by L Hôpital s rule, as well, and so we also have From this it follows that ln1 + y) y 0+ y 1 x ln 1 + 1 ) 1. x x n ln 1 + 1 ) 1 n n and so, taking exp of both sides, n x n exp1) e. Comments: a) Note how careful we had to be in our use of language. The proof is clearly discovered backwards, but we must make it clear that we are eventually) arguing forwards logically from a true statement to deduce the result we want. b) In general, x fx) L n fn) L where the first is a function it and the second is the it of a sequence), but the converse is FALSE. Do you know a continuous function f : R R which gives a counterexample to the converse? 2
f) Answer: x n n! 100) n diverges. Justification: We apply the ratio test. See the Mathematical Analysis FAQ document, Question and answer 2.1) First note that x n 0 for all n. Considering ratios of successive moduli, we have and so x n+1 x n x n+1 x n x n+1 n x n Thus the sequence diverges, by the ratio test. n + 1)!/100)n+1 n + 1 n!/100) n 100 + > 1. 3) In some cases, more than one test is acceptable: the test given here is the one that Dr Feinstein would use. If you are unsure whether or not your test would work, please ask! 1) n a) Answer: converges by the alternating series test. 2n + 3 Note: You should NOT use any form of the comparison test here. Do you know why not?) Justification: We need to check that the three conditions for the alternating series test are satisfied. Here the terms of the series are of the form 1) n 1 b n where b n 2n + 3. The required conditions are: i) b n 0 for all n [so that we have a genuine alternating series]; ii) b n b n+1 for all n; iii) b n 0 as n. These three conditions are clearly satisfied by b n 1, and so the series converges by 2n + 3 the alternating series test. Additional comments: In fact i) is redundant, because ii) and iii) together imply i). However it is NOT true that i) and ii) together imply iii), and nor is it true that i) and iii) together imply ii). Can you think of some counter-examples?) In order to avoid mistakes, I advise simply checking all three conditions. 2 b) Answer: diverges by the it comparison test. 3n + 4 Note: You should NOT use the divergence test. Do you know why not?) Justification: [We see that the terms of the series look a little like 1/n.] Set a n 2 3n+4 and b n 1 n. Then a n and b n are both positive, and a n 2 n b n n 3 + 4 n 2 3 0, ) ]0, [ 3
in our new open interval notation. [In other words n a n /b n is a strictly positive real number, and not +.] By the it comparison test, a n and b n either both converge or both diverge. Since b n is a standard divergent series, it follows that a n diverges too. c) Answer: sin n n 3 + 1 converges by the comparison test Note: You should NOT use the alternating series test. Do you know why not?) Justification: [The series can clearly be compared with an easy standard convergent series.] Set a n sin n/n 3 + 1) and b n 1/n 3. Then a n sin n n 3 + 1 b n for all n and b n is a standard convergent series. Thus a n converges by the comparison test, as claimed. d) Answer: 3 n n! converges by the ratio test. Justification: [It is clear that we should try the ratio test: see the Mathematical Analysis FAQ document, Question and answer 2.1.] Set a n 3 n /n!. For all n we have a n 0, so we can consider successive ratios of moduli of terms. We have a n+1 a n a n+1 3n+1 /n + 1)! a n 3 n /n! 3 n + 1 0 < 1 as n. Thus the series converges by the ratio test. e) Answer: 1) n 2 + 1 n diverges by the divergence test Note: You should NOT use the alternating series test, because this is a test which may prove convergence but is not designed to prove divergence. Justification: The terms of this series are a n 1)n 2 + 1. Clearly a n 1/2 as n. n In particular, the sequence of terms of the series, a n ), does not converge to 0. Thus the series diverges by the divergence test. 4) Answers The functions f and h are continuous, but g is discontinuous. Note: It is a good idea to sketch these curves to help you to see what the answers are likely to be exercise). However we require proper justification along the following lines. Justification: It is clear that each of these functions is continuous at every point except possibly at the point x 0. Thus we only need to check continuity at 0. 4
[For each function, we look at the value of the function at 0 and the it of the function as x tends to zero from the left x 0 ) and from the right x 0+ ). A function is continuous at 0 if and only if all three of these exist and are equal. Also, the two one-sided its exist and are equal if and only if the two-sided it x 0 of the function exists.] a) Here we have f0) cos 0 1, x 0 fx) x 0 x+1) 1 and x 0+ fx) x 0+ cos x 1. Since all three of these are equal, f is continuous at 0 and hence f is a continuous function from R to R by the comments above). b) Here we have g0) 0, but clearly x 0 gx) x 0+ gx) 1. Since these its are not equal to g0), g is discontinuous at 0 and hence g is a discontinuous function. [Note that we do have a two-sided it for g here: x 0 gx) 1. There is another acceptable argument which uses this fact exercise).] c) This time we have h0) 1. We consider the two-sided it sin x hx) x 0 x 0 x. This is an indeterminate form of type 0/0, so we differentiate numerator and denominator and consider cos x x 0 1 which is clearly equal to 1. Thus, by L Hôpital s rule, x 0 hx) 1 too. Since this is equal to h0), h is continuous at 0 and hence h is a continuous function from R to R by the comments above). 5