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

infinity.

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

of

**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 s_{1} 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 j_{1}. Excluding the
numbers

, there are an infinite number of rational
numbers between
and r. Let s_{2} be

one of them; then for some index j_{2} > j_{1}.
Excluding the numbers , there

is a rational number with j_{3} > j_{2}
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 N_{ε} such that

for n ≥ N_{ε}, |m − a_{n}| < ε. Since each a_{n}≤m, the condition |m − a_{n}| < ε is the
condition

m−a_{n} < ε. Because the sequence is increasing, if we find an index N_{ε}
such that ,

then m−a_{n} < ε for n≥N_{ε}, 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 a_{n} of the sequence such that m − ε < a_{n},
i.e., m − a_{n}< ε.

Denote that index n by the symbol N_{ε}. 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

formula

(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

again.

(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.