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Math 327A Exercise 2

1. The decreasing sequence converges to 0. What can you say about the sequence
{1/3, 1/2, 1, 1/6, 1/5, 1/4, 1/9, 1/8, 1/7, ...}?

Solution: The sequence is not decreasing, but its limit is still 0. If we call the sequence
Those terms go to zero as n goes to

One can also appeal to the fact that the sequence {1/3, 1/2, 1, 1/6, 1/5, 1/4, 1/9, 1/8, 1/7, ...}
is a rearrangement of the decreasing sequence which converges to 0. By the fourth
exercise of the first assignment, the rearranged sequence converges to the same limit as the
original series.

2(a). Suppose that the sequence is a list of t he rational numbers between 0 and 1.
(The sequence is a rearrangement of the sequence in part b.)

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

Solution: Let r be a real number in the range 0 < r < 1. We will find a increasing
subsequence of that converges to r. Let s1 be a rational number between r/2 and
r (There is an infinity number of such rational numbers by the first exercise set). Because
is a list of the rational numbers, for some index j1. Excluding the numbers
, there are an infinite number of rational numbers between and r. Let s2 be
one of them; then for some index j2 > j1. Excluding the numbers , there
is a rational number with j3 > j2 between and r. Continuing in that way, we
produce an increasing subsequence that 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.

Solution: Consider the sequence Those fractions are in lowest terms
except when 3 divides n. Take the subsequence of that sequence obtained by leaving out the
terms whose indices are divisible by 3. Then that subsequence is also a subsequence of the
sequence 0, 1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5.2/5, 3/5, 4/5, ..., and it converges to 2/3.

3. The ordered set of real numbers has the property that each bounded set of real numbers
has a greatest lower bound and a least upper bound.

Let be an increasing bounded sequence of real numbers, and let m be the least upper
bound of the (bounded) set of numbers in the sequence, Show that the sequence converges
to m.

Solution: We need to show that for each value ε > 0, there is an index Nε such that
for n ≥ Nε, |m − an| < ε. Since each an≤m, the condition |m − an| < ε is the condition
m−an < ε. Because the sequence is increasing, if we find an index Nε such that ,
then m−an < ε for n
Nε, automatically, i.e., when the sequence is increasing, if one term
gets within ε of m, then all subsequence terms are also within ε of m.

Because m is the least upper bound for the terms of the sequence, m − ε is not an upper
bound. Hence, there is a term an of the sequence such that m − ε < an, i.e., m − an< ε.
Denote that index n by the symbol Nε. Then, as required.

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

4(a). Compute that

Solution: Because converges to e, the subsequence also converges to
e. So, by problem 5 below (which gives the

(b). Compute

Solution: The sequence is increasing, just as the sequence that converges to e is
increasing. Hence, it has the same limit as any infinite subsequence by problem 1. Consider
the subsequence where we just use the even indices. That subsequence converges
to the limit e^2, since it is the sequence which converges to e^2, using problem 5

(c). Compute

Solution: Just as in part (b), the sequence converges to e^3

Showing that a sequence has limit m is the same as showing that the sequence
has limit 0.

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
and show that each of the summands in that expression
converges to 0 as n goes to ).

Solution: Let us show that each summand converges
to 0. Since converges to m, the sequence converges to 0. Since k
is a constant, the sequence also converges to 0. Consider the summand
converges, the set of terms in the sequence has an upper bound M.
Then where the sequence converges to 0;
hence, the sequence also converges to 0, by the preceding inequality.

6. Show that (Use problem 5).

Solution: Consider the product sequence Since the first sequence in the
product converges to e and the second sequence in the product converges to 1, the product
converges to (e)(1) = e.