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# Math 327A Exercise 2

1. The decreasing sequence
,
i.e., the decreasing sequence 1,1/2.1/3,1/4,1/5,1/6,...,1/n,...

converges to 0. The sequence {1/3, 1/2, 1, 1/6, 1/5, 1/4, 1/9, 1/8, 1/7, ...} is
not decreasing,

but it still converges to 0. By what principle from assignment 1 can you
conclude that that

sequence converges (to 0)?

**Definition:**
is a convergent sequence if it has a limiting value.

Definition: A subsequence
of a sequence
is any sequence obtained from

by eliminating some of the terms in the a_{n}-sequence. For example, the sequence of

odd positive integers is a subsequence of the sequence of all positive integers.

2.(a). Suppose that the sequence
is a list of the rational numbers between 0
and 1 (in

some order). For instance, it might be the sequence
1,1/2,1/3,2/3,1/4,3/4,1/5,2/5,3/5,4/5,...,

in which each rational is in lowest terms, and where we list the numbers with
smaller de-

nominators first.

Show that every real number between 0 and 1 is the limit
of some convergent subsequence

of . (Let r be a real number, 0 < r < 1.
Using problem 5 of assignment 1, we

know that there is a rational number between 0 and r. That number appears once
in the

list ; say it is the number
. Let be
the point half way between and r. We

know that there are an infinite number of rational numbers between
and r, i.e., an infinite

number of terms of the sequence between
and r. Let
be the first term of the

sequence after that lies between
and r. Then
Next

find a term after
that is more than half way from
to r. Then
.

Continuing in this way, you should be able to find a subsequence
of the sequence

such that .
That subsequence converges to r.)

(b). Find a subsequence of the rational number sequence 0,
1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, ...

that converges to 2/3. (The sequence
converges to 2/3. Take the

subsequence where you drop the terms whose indices n are divisible by 3. Confirm
that that

subsequence is also a subsequence of the rational number sequence given at the
beginning of

the problem.)

**Definition:** Let S be a set of real numbers. S has
an upper bound if there is an integer

which is greater than (or equal to) each number in S, i.e.,
holds for each number

.

The ordered set of real numbers has the property that each
set of real numbers with an

upper bound has a least upper bound, and each set of real numbers with lower
bound has a

greatest upper bound.

3. Let be an
increasing sequence of real numbers that has an upper bound, i.e., the

set of numbers occurring in the sequence has an upper bound, and let m be the
least upper

bound of the set of numbers in the sequence. Show that the sequence converges to
m.

It is a fact that .
Using that fact, show the facts given in exercise 4

below.

4(a). Compute the equality

(b). Compute

(c). Compute

Showing that a sequence
has limit m is the same as showing that the sequence

has limit 0. (See exercise 1 of assignment
1.)

5. Suppose that
converges to m and that converges to k. Show
that

converges to mk. (Hint: Show that
converges to 0. To do that,

write mk − as
and show that the summands (m −
)k and

determine sequences
that converge to 0.

(In dealing with the second sequence, you may assume the fact (or prove the
fact) that if a

sequence consists of a set of numbers that
has an upper bound and a lower bound

(that property holds when the sequence converges, for instance) and if a
sequence

converges to 0, then the sequence converges
to zero.)

6. Show that (Use problem 5).