Exam (elaborations)
MAT2613 - EXAM PACK (2022)
- Institution
- University Of South Africa (Unisa)
Contains frequently asked Questions and Answers from Past Exam Papers and Assignments. Also Includes a Summary of important Study Notes.
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MAT2613EXAM PACK
Revision PACK
Questions. Answers
, 3
I OCTOBER/NOVEMBER )0/7EXAMINATION PAPER AND MEMORANDUM I
QUESTION l
1.1 Use a proof by contradiction to prove that the following statement is true.
2n ;::: 2n for all positive integers n.
[Hint: You may assume the well ordering axiom: Every non-empty set of positive integers has a least
Open Rubric
~~] 00
SOLUTION
Contradiction: There exist at least one positive integer m such that 2m < 2m.
Assumption false for m = 1 and m = 2. The statement must then be: There exist at least one positive integer
m > 2 such that 2m <2m.
Let M = (m lm > 2, m EN, 2m < 2m}. This set M bas a least element by the well·ordening axiom.
Let mo be this element. Then mo > 2 and 2mo < 2m 0 (1)
However, mo - 1 < mo and mo- 1 ¢ M, so 2<mo-l) ;::: 2 (mo- 1) (2)
and so from (1) and (2) we have since (2) is 2mo ;::: 4m 0 - 4 that 4m0 - 4 ~ 2mo < 2m0 , i.e 2m0 < 4 or m0 < 2
which is a contradiction.
1.2 Give the contrapositive of the following statement:
00
If L, Or is convergent then (an) is a null sequence. (2)
rei
[10]
SOLUTION
00
If (an) is not a null sequence then L:ar is divergent.
r•l
QUESTION%
Let (an) be the sequence of real numbers defined by a 1 = I and an+l = ,JiCi;, for n EN.
Show that (an) converges and find the limit.
[Hint: Show that 1 ~an < an+l < 2 for all n EN using mathematical induction.] f81
SOLUTION
a1 = 1 and On+ 1 = -J24, 'if n .
Following the hint we have to prove that 1 ~ an < an+2 < 2 'if n. (*)
For n = 1 we have a 1 = 1 ami a 2 = ,J2 thus (*) is true for n = I.
Suppose(*) is true for n = k, i.e 1 ~ at < ak+l < 2 (**)
Then we have from(**) that 2 ~ 2ak < 2aA:+1 < 4 so that ,J2 ~ ,J2iii < ~ < 2.
t
Open Rubric
, 4
But
A - ak+l and J2ak+l = ak+2
so 1 < .J2 ~ ak+l < ak+2 < 2 and the equation (**)is true.
We thus have an increasing sequence which is bounded above by 2.
Suppose
lim an
n-too
= L. Then also11--tOO
lim an+ I = L
We have
lim an+ 1 lim .J2ci:, = Jlim 2an
= n-too
11--too n-too
L =.fi-JI i.e -Jl = v'2 or L =2.
QUESTION3
Prove from first principles that the sequence (an) with
2n2 +5
a1 = 0, an = ., when n ?: 2
n-- 1
converges. (7)
SOLUTION
. . 2n 2 + 5 2 + 2.. /
We suspect that lrm an
n-too
= lun
11--tOO n 2 - 1
= lim ~ =2
n-too ( - ~
'-"
Let c > 0 be given. For n :;::: 2 we have
Since
> n when n :;::: 2 we have
lan- 21
7 7/
- - - < - for n > 2
n -l-n
2 -
Clearly
7 7
-<e~n>
n e
By the Archimedean principle there exists ; : : N with N > ~.
f:
For such an N e N we have
• 7 7
n 2: N => n > - => lan - 21 < - < c
e n
, 5
Since c > 0 was arbitrary we have lim a,
n~oc
= 2.
QUESTION 4
4.1 Test each of the following series for absolute convergence, conditional convergence or divergence:
r'
4.1.1
oo
I:
r=l
<- 1r ·
3r + 3r1
(4)
SOLUTION
. Ia, I
1un
11--HXl
= I'lm
ri~OO-
3" rl
+V=
.3 '
~
1·un
IJ~OO-
~+
-3,--
1
1
= 3- V/
3n! 3JJ!
3
since ( ; \is a null sequence
3nJ
lim a, '::f= 0 and fr~e contrapositive of the vanishing condition f (-1) a, is divergent.
Since lim
-00 lanl '::f= 0, -00 ~
00 1
4. t. 2 L
r=l
<- 1y ~~Y'2r
r::;;=====
1 2 -
(7)
SOLUTION
1
Let lar I = ~~r:==:<===
.:/2r2- 1
1 1 1
We have ~
2
> 3r::;-;; =- -
1 -2
Y'2r 1 -v2r·/' r J
By the p- test
-
p = -2 < 1 the series
00
L 2
1 /
diverges and hence 1
1 00
L
1 / ~~.l
diverges so that ~
diverges.
3 r=lr3 23 r=ol r' 2 3
r•l 2r 2 - 1
Forconditionalllyconvergence: We have lim
r---+00 v
lari.=.P( A1so2(r + 1)2 - 1 > 2r 2 -1 and thus
so that the series f
r=l
larl is decreasing. We thus have that the given series f
r=l
(-1Y
1
J2r 2 - 1
is conditiondtt{
convergent:
[Iff (x} = (2x 2 - 1r 3
I
then f 1 (x)
}
= -- (2x 2 -
3
1r'.
4
4x
4
= -- (2x 2 -
3
1r 4
3
< owhich shows that the series
L larl is decreasing]
4.1.3 L (-1Y
00
r=2
(1
r2
sin- 7r)
r
(7)
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