ENGR 1405 Engineering Mathematics 1
Faculty of Engineering and
Applied Science
2000 Fall
Problem Set 5
Solutions
����������� In questions 1-6 discuss convergence
of the sequence.���
����������� If it is
not given, find the general term.
1.�����
�������� 
The numerators are increasing by 4 with each term
��� presence of� 4n� in the numerator of the general term.
The denominators are increasing by 3 with each term
��� presence of� 3n� in the denominator of the general term.
If� n� starts at 1, then
An alternative is
Therefore this sequence converges to 4/3.
2.�����
�������� 
�
Multiplying the numerator and denominator of all �even terms� (terms an with n even) by 2, the sequence is
We can now see that
or
Therefore this sequence converges to 1.
3.�����
�������� 
��
The denominators are the square roots of consecutive even integers.
With� n� starting at 1,
An alternative is
Clearly this sequence converges to 0.
4. ����
�������� 
�
Let� bn = ln an .�� Then
Therefore this sequence converges to 8.
5.� ���
�������� 
�
Therefore this sequence converges to 0.
6.�����
�������� 
�
The limit contains an (� - �) type of indeterminacy.
Rationalizing the numerator:
Therefore this sequence converges to 0.
7.����� Every hour each virus that is at least two
hours old creates one new copy of itself.
����������� A
newly created copy becomes productive in the same way two hours later.
����������� Assume
that all of these viruses are immortal (never die).
����������� At
hour 0 a single virus is created and placed on a petri dish.
����������� At
hour 1 only that single virus is present on the petri dish.
����������� At
hour 2 there are two viruses present (the original and its newly created copy).
��� (a)�� Write down the next seven terms in the
sequence { fn } of the number of viruses present at hour n:
����������������������� {
1, 1, 2,��� ,��� ,��� ,��� ,���
,��� ,��� , ... }
[Note:�� this
sequence is known as the Fibonacci sequence.]
At hour 3
the original virus reproduces again but its first offspring does not, for a
total of 3.
At hour 4
both viruses that were present at hour 2 each produce one new virus, to be
added to all the viruses present at hour 3, for a total of 5.
At each
subsequent hour n, all viruses present at hour n - 2 each produce one new virus, to be added to all the
viruses present at hour� n - 1.
The
recursion relationship is therefore
an� =� an-1 + an-2
The first ten terms of the Fibonacci sequence are
{ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ... }
��� (b)�� Now suppose that each virus becomes
productive when only one hour old.��
Write down the first four terms and the general term in the sequence { gn
} of the number of viruses present at hour n.
At hour 0
a single virus is created and placed on a petri dish.
At hour 1
that single virus and its copy are present on the petri dish.
At hour 2
each of the two viruses produces another copy, for a total of 4.
Clearly
the number of viruses doubles each hour, to produce the geometric sequence
{ 1, 2, 4, 8, ... , 2n, ... }
��� (c)�� Is the sequence� { fn }�
convergent?
NO
The sequence clearly diverges (to
infinity).
[This can be seen from the recursion relation;
each successive term is the sum of two previous positive integer terms.
The sequence is therefore an increasing sequence of integer values,
which must, therefore, diverge to infinity.]
��� (d)�� Is the sequence� { gn }�
convergent?
NO
The sequence clearly diverges (to
infinity).
[It is the standard sequence { rn }
with r > 1.]
��
In questions 8-10, test the series for convergence.�� If it converges, then find its sum.
8.�����
�������� 
�
�is a convergent
geometric series 
�with sum
�is a convergent
geometric series 
�with sum
The sum of two convergent geometric series is a convergent series.
Therefore the series is convergent with sum
9.�����
�������� 
�
One could apply the limit comparison
test with reference series 
to establish that the series above does converge, but its sum is also required.
It is clearly not a geometric series.�� Try to write it as a telescoping series.
Partial fractions, using the cover-up rule:
���� 
nth partial sum (with n starting at 2, so that s2 = a2 and s1 is undefined):
This is indeed a telescoping series.
Therefore the series converges to 3/2.
10.���
�������� 
�
Divergence test (nth term test):
�
Therefore this series is divergent.
11.��� Write the number 
as a ratio of two relatively prime integers.
[Note:� the
answer to this question is an ancient approximation to the value of
�������� p� = 3.14159265... ]
3.142857... �=� 3 + 142857�10-6 + 142857�10-12 + 142857�10-18 + ...
= 3 + Sum of geometric series (a = 142857�10-6,� r = 10-6)
| r | < 1��� ��� series converges to�
a / (1 - r)� =� 142857 / 999999� =� 15873 / 111111�
=� 5291 / 37037� =� 143 / 1001� =� 1 / 7
[or just check that 999999 is an integer multiple of 142857]
Therefore�� 3.142857... �= �3 + (1/7)� = �22/7 .
[Notes on divisibility (- the rules for divisibility by 9 and by 3 were used above):
An integer is divisible by 2 if its last digit is even.
An integer is divisible by 3 if the sum of its digits is an integer multiple of 3.
An integer is divisible by 4 if its last two digits form a multiple of 4.
An integer is divisible by 5 if its last digit is 0 or 5.
An integer is divisible by 6 if it is even and the sum of its digits is an integer multiple of 3.
An integer is divisible by 8 if its last three digits form a multiple of 8.
An integer is divisible by 9 if the sum of its digits is an integer multiple of 9.]
12.��� Every 160 microseconds (= 1.6�10-4 seconds) half of the remaining amount of the C' isotope of
the element radium disintegrates radioactively into polonium.�� At time t = 0 just over a kilogramme
of radium C' is alone in a sealed box.��
After 160 microseconds 500 grammes of polonium has been produced.�� At� t
= 320 microseconds another 250 grammes of polonium has been produced, making
750 grammes in total.�� At� t = 480 microseconds another 125
grammes of polonium has been produced, making 875 grammes in total, and so on.
(a)������ Find
the amount of polonium produced after 160n microseconds, where n
is a positive integer.
Let� an� =� amount produced in the 160 microsecond interval ending at time t = 160n,
then� an� =� 1000 / 2n
The total produced by time t = 160n is
��
which is a geometric series,�� a = 500,� r = 1/2.�
Therefore
��
(b)������ Find
the total amount of polonium produced in a day, correct to the nearest gramme.
One day = 24 � 3 600 s �=� 86 400 s = 86 400 000 000 m s
��� n� =� 86 400 000 000 / 160� =� 540 000 000
��� (1/2)n� is extremely close to zero.
Therefore, correct to the nearest
gramme, the total amount of polonium produced in a
day is
s = 1000 �g
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