Math 143, Calculus III, Fall 2013
Written Homework Example
1. Write up the student study guide solution to 12.1.62
2n − 3
Determine whether the sequence an =
is increasing, decreasing, or not monotonic. Is the sequence
3n + 4
bounded?
It turns out that {an } is an increasing sequence and there are many ways to show this.
2(1) − 3
1
1
2(2) − 3
First we observe that it is not decreasing since a1 =
=− <
=
= a2 . So it is either
3(1) + 4
7
10
3(2) + 4
increasing or not monotonic.
Now consider the definition. We know that {an } is increasing if an < an+1 for all n ≥ 1. In this instance,
an
=
an+1
=
2n − 3
, and
3n + 4
2(n + 1) − 3
2n − 1
=
.
3(n + 1) + 4
3n + 7
So we need to show that
2n − 1
2n − 3
<
= an+1
3n + 4
3n + 7
for every n ∈ N. That is, we need to show that
an =
(2n − 3)(3n + 7) < (2n − 1)(3n + 4),
or foiling (it’s safe, it’s easy, it’s fun) that
6n2 + 5n − 21 < 6n2 + 5n − 4.
This last inequality follows since
−21 < −4.
We conclude that
an =
2n − 3
2n − 1
<
= an+1
3n + 4
3n + 7
for every n ∈ N as required.
2n − 3
is increasing is by relating it to a function. Recall that a function
3n + 4
g(x) said to be increasing if g(x1 ) < g(x2 ) for all x1 < x2 in its domain. Since {an } lives on the function
2x − 3
(that is, an = f (n) for all n ∈ N) and N is in the domain of f (only x = −4/3 is
f (x) =
3x + 4
troublesome), it follows that if f is increasing, then an = f (n) < f (n + 1) = an+1 for all n ∈ N, that is
that {an } is increasing. So we are done if we can show that f is increasing, which means that we can bring
more weapons to this fight. In Calculus I, we learned that if f 0 (x) > 0 on [0, ∞), then f is increasing
17
there. Since f 0 (x) =
(using the quotient rule indeed!) and this function is obviously positive
(3x + 4)2
for all x ≥ 1, we have that f is increasing and thus {an } is an increasing sequence.
Another way to show that an =
To show that this sequence is bounded, we need to show that it is bounded above and below, that is, that
there are natural numbers N and M such that N ≤ an ≤ M for all n ∈ N. Since we have already shown
that {an } is increasing, it follows that a1 ≤ an for all n ∈ N, and thus we may take −1/7 = a1 to be N .
Now theorem 3 in the text tells us that if lim f (x) = L and f (n) = an for all n ∈ N, then lim an = L,
x→∞
n→∞
2x − 3
2
2
and L’Hospital’s rule gives lim f (x) = lim
= lim
= , so we know that lim an = 2/3. It
x→∞
x→∞ 3x + 4
x→∞ 3
n→∞
3
turns out that this is enough to show that {an } is bounded. We can proceed directly or use definition 2.
Even without definition 2, however, and just using our work-a-day definition of limit, it easy to believe
that {an } is bounded. After all, how could a sequence be unbounded if it converges? How could it be
at various times arbitrarily large if it is getting closer and closer to 2/3? In any event, let’s show that
2n − 1
< 2/3 for all n ∈ N, that
an < 2/3 for all n ∈ N. To do so, it is enough to demonstrate that an =
3n + 4
is, that 3(2n − 1) < 2(3n + 4), or 6n − 3 < 6n + 8, or −3 < 8, and of course the last inequality is obvious.
To show that an → 2/3 implies {an } is bounded using definition 2 is a bit hairy for a math 143, but for
the record one proceeds as follows. Definition 2 states that a sequence lim an = L if for each > 0 there
n→∞
is N ∈ N such that |an − L| < for all n > N . So, given = 1/3, there is N ∈ N such that
|an − 2/3| < 1/3
for all n > N , that is
−1/3 < an − 2/3 < 1/3
or
1/3 < an < 1
for all n > N . It follows immediately that {an } is bounded above by the maximum value of 1, a1 , . . . , aN
and below by the minimum of 1/3, a1 , . . . , aN .
2. Write up the student study guide solution to 12.2.58a
A certain ball has the property that each time it falls from a height h onto a hard, level surface, it
rebounds to a height rh, where 0 < r < 1. Suppose that the ball is dropped from an initial height of H
meters.
(a) Assuming that the ball continues to bounce indefinitely, find the total distance that it travels. (Use
1
the fact that the ball falls gt2 meters in t seconds).
2
The hint about how fast the ball drops really belongs with part (b) of this problem. To find distance
travelled, note that each rebound is r times the previous. That is, we drop the ball from height H (so
it travels distance H), then it rebounds to height rH (so travels rH) and drops from rH (travelling
rH again), then rebounds again to r(rH) and drops from r2 H, and so on indefinitely. We can think
of the distance as the sum of the distances travelled on each bounce (from the floor to the top to the
floor again. So the total distance the ball travels (if it is finite) is
(0 + H) + (rH + rH) + (r2 H + r2 H) + (r3 H + r3 H) + · · ·
(H) + (2rH) + (2r2 H) + (2r3 H) + · · ·
Of course,
2rH + 2r2 H + · · · =
∞
X
2rHrn−1 =
n=1
2rH
1−r
by the Geometric Series Test so
= H + 2rH + 2r2 H + · · · = H +
is the total distance travelled.
2rH
1+r
=
H
1−r
1−r
3. Write up the student study guide solution to 12.2.70
X
X
X
If
an and
bn are both divergent, is
(an + bn ) necessarily divergent?
The answer is no. Consider the following example: let {an } be the sequence with an = 1 for all n ∈ N
and {bn } be the sequence with bn = −1 for all n ∈ N.
X
X
Recall the Test for Divergence: if lim cn 6= 0 for a sequence {cn }, then
cn diverges. Thus both
an
n→∞
X
and
bn diverge since (by the limit laws on page 714)
lim an = lim 1 = 1 6= 0
n→∞
n→∞
and
lim bn = lim −1 = −1 6= 0.
n→∞
n→∞
X
X
X
But
(an + bn ) =
(1 − 1) =
0, and this converges to 0. Of course this last assertion should be
∞
justified: note that the sequence {an + bn }∞
n=1 = {0}n=1 so that the associated sequence of partial sums
is 0, 0 + 0, 0 + 0 + 0, . . . , that is, the sequence of partial sums is also {0}∞
n=1 which converges to 0 (again
by the limit law on page 714).
X
X
X
It is worth pointing out that theorem 8 says: If
an and
bn are convergent, then so is
(an + bn ).
X
Theorem 8 does not say that the converse is true, that is, Theorem 8 does not imply that if
(an + bn )
X
X
converges then both
an and
bn do as well (we just showed that this is not so).