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.

3. Show all of your work. Your work must justify the answer that you give.

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


(a)   

n = 1 
(-1)n+1 n3
2n
          (b)   

n = 1 
(-1)n+1
2n+3

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


an+1
an
= (n+1)3
2n+1
n3
2n
= 1
2

n+1
n

3
 
.



lim
n  
an+1
an
= 1
2

lim
n 

n+1
n

3
 
= 1/2 < 1.

Hence, n = 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  
an
bn
=
lim
n  
n
2n+3
= 1/2 < 1.

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

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



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

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 (-, ). 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- 1
3!
x3+ 1
5!
x5- 1
7!
x7+...,        x (-, ).




File translated from TEX by TTH, version 2.79.
On 12 May 2001, 09:25.