Math 142 Homework # 9 – Solutions
1–7. Determine whether the sequence converges or diverges. If it converges, find the
limit.
1. an = cos(nπ/2). We have
(an ) = (1, 0, −1, 0, 1, 0, −1, 0, 1, 0, −1, . . . ).
In other words an = 0 for even n, and an = ±1 for odd n. Therefore the sequence (an ) is
not convergent.
n cos n
. Since
2. an = 2
n +1
n
n
6 an 6 2
,
− 2
n +1
n +1
we can use the
Theorem to deduce that lim an = 0.
Squeeze
1
3. an = n sin
. Let
n
sin(1/x)
.
f (x) =
1/x
Using l’Hospital’s Rule we get:
− cos(1/x) · x12
1
lim f (x) = lim
= lim cos
= 1.
1
x→∞
x→∞
x→∞
x
− x2
4. an =
4n − 3
.
3n + 4
lim an = lim
4−
3+
3
n
4
n
4
= .
3
5. an = arctan n.
lim an = lim arctan x =
x→∞
π
.
2
6. an = ln(n + 1) − ln(2n). By the continuity of ln:
n+1
n+1
1
lim an = lim ln
= ln lim
= ln = − ln 2.
2n
2n
2
√
7. an = n.
√
lim an = lim x = ∞.
x→∞
1
8. A sequence {an } is given by a1 = 2, an+1 = 3−a
. (a) Prove that {an } is decreasing
n
and bounded from below. Deduce that limn→∞ an exists. (b) Find limn→∞ an .
(a) Let us prove first that for all n > 1,
an+1 < an .
(*)
It is easy to check that (*) is true for n = 1:
a1 = 2,
a2 =
1
= 1,
3−2
a2 < a1 .
Now assume that (*) holds for some n > 1. Then
an+1
3 − an+1
1
3 − an+1
an+2
< an
> 3 − an
1
<
3 − an
< an+1 .
By induction (*) is true for all n > 1. We will prove that all terms of this sequence are
positive (and therefore they are bounded from below by 0). We need to prove:
an > 0
(**)
for all n > 1. It is obvious that (**) is true for n = 1 (because a1 = 2 > 0). Now suppose
that (**) is true for some n > 1. Then
an > 0
3 − an < 3
1
1
> >0
3 − an
3
1
an+1 > > 0.
3
By induction we conclude that (an ) is bounded from below by 0. Since (an ) is decreasing
and bounded from below, it follows that lim an exists.
(b) Let g = lim an . Then
1
3−g
3
g − 3g + 1 = 0
√
3± 5
g=
.
2
We have obtained two solutions for g, however only one of these solutions represents the
limit. Since an is a decreasing sequence, a1 is its biggest term, and so
g=
an 6 2
√
for all n > 1.
√
But 3+2 5 > 3+2
= 2.5. Therefore 3+2 5 cannot be the limit of an (all an ’s stay below 2
2
and so cannot approach this number which is above 2.5). Therefore we have
√
3− 5
lim an =
.
2
9. Fibonacci sequence (fn )∞
n=1 is a sequence of natural numbers defined by a recursive
formula:
f1 = f2 = 1,
fn+2 = fn+1 + fn ,
for all n > 1.
(a) Write down the first 10 terms of this sequence. (b) Prove by induction the following
formula:
√
√
(1 + 5)n − (1 − 5)n
√
fn =
.
(***)
2n 5
(c) Find
lim
fn
.
fn−1
(a)
1, 1, 2, 3, 5, 8, 13, 21, 34, 55.
(b) It is easy to see that (***) is true for n = 1 and n = 2. Suppose now that (***) is
true for some n and n + 1. We will prove that then (***) must be also true for n + 2. To
simplify computations let us use the following simple identities:
√ 2 1 √
√
1
1+ 5 =
1+2 5+5 =3+ 5
2
2
√ 2 1 √
√
1
1− 5 =
1 − 2 5 + 5 = 3 − 5.
2
2
Then we have
fn+2
√
√
5)n + (1 + 5)n+1 − (1 − 5)n+1
√
= fn + fn+1 =
2n+1 5
√
√
√
√
(1 + 5)n (3 + 5) − (1 − 5)n (3 − 5)
√
=
2n+1 5
√
√
√
√
(1 + 5)n · 12 (1 + 5)2 − (1 − 5)n · 21 (1 − 5)2
√
=
2n+1 5
√
√
(1 + 5)n+2 − (1 − 5)n+2
√
.
=
2n+2 5
2(1 +
√
5)n − 2(1 −
√
Therefore the formula (***) is true for n + 2. By induction (***) is true for all n > 1.
(c) Using (b) this part is not difficult. We have
lim
fn
fn−1
√
√
√
√ n
(1− 5)n
√
(1
+
5)
−
1 (1 + 5) − (1 − 5)
1
(1+ 5)n−1
√
√
√
.
= lim
= lim
n−1
2 (1 + 5)n−1 − (1 − 5)n−1
2
1 − (1−√5)n−1
√
n
(1+ 5)
√
√
√
Now lim(1+ 5)n = ∞ (because 1+ 5 > 1), and lim(1− 5)n = 0 (because |1− 5| < 1).
Therefore
√
fn
1+ 5
lim
=
.
fn−1
2
8–9. Determine if the following series is convergent or divergent. If convergent, find its
sum.
∞
X
1
9.
. This is a geometric series with a =
2n
e
n=1
1
e2
and r =
1
.
e2
Therefore its sum is:
1/e2
1
.
1 = 2
e −1
1 − e2
10.
∞
X
3−n 4n . This is a geometric series with a =
n=1
because r > 1.
4
3
and r =
4
,
3
and it is divergent