This exercise sheet contains some additional questions that may be useful for revision or review of the material in these web pages.
This is work in progress. Please return here later.
Exercise.
Using the theorem on subsequences, and/or the theorem on boundedness
of convergent sequences, and/or the theorem bounding the limit of a bounded
sequence, prove the following sequences (
)
with th term
as given do not converge.
(a)
=0
if is prime,
=1
otherwise.
(d)
=
. (Hint: consider the
subsequences
()
and
()
where ()=2
+
2
and ()=2
+32
.)
Exercise.
Prove that ()
()
2
for sufficiently large . (Take logs of both sides.
What properties of the function do you require?)
Hence show that the series
=2
1
()
()
converges.
Exercise.
A function ()
is defined by setting ()
=
lim
-
+
.
Prove that ()
is defined for all
(i.e., the limit above exists for all
)
and takes the following values:
()=
if
1
;
and
()=-
if
1
.
(Hint: Split into cases and divide top and bottom of the
fraction by appropriate numbers. You may find it helpful
to use that fact that
0
if
1
.)
Solution to Exercise 2, in the special case when 0
1
:
Note that for these values of we have
-
+
=
1
-1
1
+1
0-10+1
=- as
by standard null sequences
, 1
and continuity of the arithmetic
operations. Thus ()=-
for 0
1
.
The next two exercises concern some common misconceptions and
errors in beginning analysis. Many of these examples are worth learning.
Some are easy but some are rather subtle and tricky. (One or two are
dreadful howlers that I hope no-one reading this would ever consider
writing.) If you have a good feel
for things like this you are
well on the way to being rather good at analysis.
Exercise.
The following statements are all false.
In each case, find a counterexample.
(a) If the sequence (
)
diverges and the sequence (
)
diverges then their sum (
+
)
diverges.
(b) If the sequence (
)
diverges and the sequence (
)
diverges then their product (
)
diverges.
(c) If the sequence (
)
is a null sequence
then
=1
converges.
(d) If
for all
and
as
then
.
(e) If (
)
is a bounded sequence and has
a convergent subsequence
()
as
then
.
(f) If
=1
=
and
=1
=
both converge, then
=1
(
)
converges and equals
.
(g) If (
)
is a null sequence then
0
as
.
(h) If the sequence (
)
does not converge
then 1
0
as
.
(i) If (
)
and (
)
are both sequences
and
0
as
then either
0
or
0
as
.
(j) If
0
as
and
=1
converges, then
=1
converges. (Hint: consider an appropriate conditionally convergent
series.)
Solution to Exercise 3(a):
Let the sequence (
)
be defined by
=(-1)
and
(
)
be defined by
=(-1)
+1
.
Then both these sequences diverge, and yet
their sum is
+
=-1+1
or 1+-1
which defines the constant sequence with value 0, which converges.
Sample solutions to selected parts
Solution to 1(a): Let (
)
be the subsequence of (
)
consisting of all those
with prime, and let (
)
be the subsequence of (
)
consisting of all those
with non-prime. Both (
)
and (
)
are valid subsequences as there are
both infinitely many primes in and
infinitely many non-primes in . But (
)
is the constant sequence with value 0 and (
)
is the constant sequence with value 1.
Therefore (
)
does not converge, for if
as
then
and
by the theorem on subsequences, so =0 by the
theorem on uniqueness of limits as
0
since constant sequences converge to their value, and
=1 by the theorem on uniqueness of limits as
1
as this is a constant sequence also.
But this means 0=1, which is false.