Exam 3 Exam 3                              Friday, May 11, 2001

NAME:

Student Number:------            No.----           Marks ----

Instructions:

1. You have 30 minutes for this exam. You are not allowed to use any calculators.

2. This exam contains 2 questions and has a total of 100 marks.

1. (60 Marks)    Determine whether each of the following series converges absolutely, converges conditionally or diverges.

 (a) ¥å n = 1 (-1)n+1 n32n (b) ¥å n = 1 (-1)n+12n+3

Solution. (a) Let an = [(n3)/(2n)]. Than an+1 = [((n+1)3)/(2n+1)] and

 an+1an = (n+1)32n+1 n32n = 12 æè n+1n öø 3 .

 lim n® ¥ an+1an = 12 lim n®¥ æè n+1n öø 3 = 1/2 < 1.

Hence, ån = 1¥(-1)n+1[(n3)/(2n)] converges absolutely.

(b) Let an = |[((-1)n+1)/(2n+3)]| = [1/(2n+3)]. Let bn = 1/n. Then

 lim n® ¥ anbn = lim n® ¥ n2n+3 = 1/2 < 1.

Since ån = 1¥[1/n] diverges, it follows that ån = 1¥[1/(2n+3)] diverges. Note that ån = 1¥[((-1)n+1)/(2n+3)] is an alternative series and satisfies (i) an ³ an+1 for n Î \mathbb N and (ii) limn® ¥an = 0.

It follows that ån = 1¥[((-1)n+1)/(2n+3)] converges conditionally.

2. (40 Marks). Find the Maclaurin series for f(x) = sinx, x Î (-¥,¥) and show that it represents sinx for all x Î (-¥, ¥). (You may use limn® ¥[(|x|n+1)/((n+1)!)] = 0 for each x Î (-¥, ¥)).

Solution: f(x) = sinx,       f(0) = 0.

f¢(x) = cosx,        f¢(0) = 1.

f¢¢(x) = -sinx,        f¢¢(0) = 0.

f(3)(x) = -cosx,        f(3)(0) = -1,

f(4)(x) = sinx,        f(4)(0) = 0,

..........................................

Since |f(n+1)(x)| = |cosx| or |sinx|, |f(n+1)(c)| £ 1 for all c Î (-¥, ¥). It follows that

 |Rn(x)| = |f(n+1)(c)|(n+1)!) |x|n+1 £ 1(n+1)!) |x|n+1.

Since limn® ¥[(|x|n+1)/((n+1)!)] = 0, it follows that limn® ¥|Rn(x)| = 0. Hence,

 sinx = x- 13! x3+ 15! x5- 17! x7+...,        x Î (-¥, ¥).

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On 12 May 2001, 09:25.