Brian E. Veitch
5
Sequences and Series
5.1
Sequences
A sequence is a list of numbers in a definite order.
• a1 is the first term
• a2 is the second term
• an is the n-th term
The sequence {a1 , a2 , a3 , ..., an , ..., } is a sequence and we denote it by
{an } or {an }|∞
n=1
Examples of Sequences
n2
n2
1)
a
=
n
n2 + 1
n2 + 1
n
n
1
1
2)
−
(n + 3)
an = −
(n + 3)
2
2
2π
2π
3)
cos
an = cos
n
n
n2
1 4 9 16
, , , , ..., 2
, ...
n +1
2 5 10 17
4 5 6 7
8
, ,
, ,
, ...
( −2 4 −8 16 −32
)
√
1
1
2
1, −1, − , 0, cos(2π/5), , cos(2π/7),
, ...
2
2
2
Some sequences aren’t defined so clearly. Here’s an example of a sequence that’s defined
recursively. Recursively means the next term in the sequence is determined by previous
terms. The next sequence is probably one of the more famousj sequences.
Example: Fibonacci Sequence
{fn } defined recursively
1. f1 = 1
2. f2 = 1
3. f3 = 2
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Brian E. Veitch
4. f4 = 3
5. f5 = 5
6. f6 = 8
7. fn = fn−1 + fn−2
The next term in the Fibonacci Sequence is the sum of the previous two terms.
Definition 5.1 (Limit of a Sequence). A sequence {an } has the limit L and we write
lim an = L or simpliy an → L as n → ∞ if we can make the terms of aN as close to L as
n→∞
we make n sufficiently large.
If limn→∞ an exists, we say it converges. Otherwise, we say it diverges.
Definition 5.2 (The Formal Definition of Convergence). We will not really use this definition. It is, however, good to know.
A sequence {an } has a limit L if for every > 0, there is a corresponding integer N such
that if n > N , then |an − L| < This basically says, If you believe the limit of this sequeence is L, then find me the N th
term such that every term after aN is within of L.
Example 5.1. Let an =
1
. I claim the limit is 0, i.e., an → 0 as n → ∞.
n
Let = .1. My job is to find the term where an is finally within 0.1 of L = 0.
an =
1
= 0.1
N
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Brian E. Veitch
1
=N
0.1
N = 10
So after N = 10, every term of an is within 0.1 of L = 0.
1 1
1 1 1 1 1 1
1, , , ..., , , , , , , ...,
2 3
9 10 11 12 13 14
Do you see how after the 10-th term, an is within 0.1 of L = 0?
Definition 5.3 (Divergence). lim = ∞, means that for every positive number M , there is
n→∞
an integer N such that if n > N , then an > M .
This definition basically says, ”‘If you claim the limit is ∞, then I’ll give you a very large
number (like 100,000,000). You need to tell me the term in an where an > 100, 000, 000. By
doing this for any extremely large number, you essentiall prove limn→∞ an = ∞
Let’s talk Limit Laws
If an and bn are convergent sequences and c is some constant:
1. lim an ± bn = lim an ± lim bn
n→∞
n→∞
n→∞
2. lim c · an = c lim an
n→∞
n→∞
3. lim an · bn = lim an · lim bn
n→∞
n→∞
n→∞
an
limn→∞ an
=
if lim 6= 0
n→∞ bn
n→∞
limn→∞ bn
p
5. lim (an )p = lim an
4. lim
n→∞
n→∞
Before we do some examples, we have two very useful theorems.
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Brian E. Veitch
Sequence Theorems
Theorem 5.1 (The Squeeze Theorem). If an ≤ bn ≤ cn for n ≥ N and lim an = lim cn =
n→∞
n→∞
L, then
lim bn = L
n→∞
Another (not named) theorem
If lim |an | = 0, then lim an = 0
n→∞
n→∞
Ok, ok.. now some Examples
Example 5.2. Write out the first five terms of an =
2n
+1
n2
2 4 6 8 10
, , , , , ...
2 5 10 17 26
It appears the sequence converges to 0, an → 0.
Example 5.3. Write out the first five terms of an =
(−1)n
(n + 1)!
Recall that n! = n · (n − 1) · (n − 2) · (n − 3) · ... · 3 · 2 · 1
−1 1 −1 1 −1
, ,
,
,
, ...
2 6 25 120 720
This is called an alternating sequence since the signs alternate between −1 and 1. It also
appears the sequence converges to 0.
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Brian E. Veitch
Example 5.4. Write out the first five terms for the sequence that’s defined as a1 = 6, an+1 =
an
n
Note that n = 1 doesn’t mean a1 , n = 2 doesn’t mean a2 .
• a1 = 1
• a2 = a1+1 =
a1
1
= =1
1
1
• a3 = a2+1 =
1
a2
=
2
2
• a4 = a3+1 =
a3
1/2
1
=
=
3
3
6
• a5 = a4+1 =
a4
1
=
4
24
n
n→∞ n + 1
Example 5.5. Find lim
Most of these can be done using methods we already know.
n
=
n→∞ n + 1
lim
1
lim
n→∞
1+
1
n
1
1+0
= 1
=
Just to verify this, let’s look at the first few terms and see if it appears the sequence
approaches 1.
1 2 3 4 5 7
99
99999
, , , , , , ...,
, ...,
, ...
2 3 4 5 6 8
100
100000
ln n
n→∞ n
Example 5.6. Find lim
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5.1 Sequences
Brian E. Veitch
Ooooh, I think we have a rule that helps with this. What was it called...L’Hospital’s Rule!
ln n LH
1/n
→ lim
n→∞ n
n→∞ 1
= 0
lim
Example 5.7. Is an = (−2)n−1 convergent or divergent?
{1, −2, 4, −8, 16, −32, 64, −128, 256, ...}
It appears an is diverging. The numbers keep getting larger and larger. They are also
alternating. Some of the terms appear to approach ∞ while the others −∞. This really just
adds more to why this sequence diverges.
Example 5.8. Does an =
(−1)n
converge?
n
Yes. If you look at the sequence
1 −1 1 −1 1
1 −1
−1, ,
, ,
, , ...,
,
, ...
2 3 4 5 6
100 101
It appears to converge to 0. So how can we prove that? Let’s use our un-named theorem from a couple of pages ago.
(−1)n 1
=
|an | = n n
We know
1
→ 0 as n → ∞. Therefore, since |an | = 0, an must converge to 0.
n
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5.1 Sequences
Brian E. Veitch
Example 5.9. Let an = cos(π/n)
lim cos(π/n) = cos(0) = 1
n→∞
Example 5.10. Where is an = rn , r is a constant, convergent?
Through a little work, which I’ll show you in class, we have
lim rn =
n→∞


0, if − 1 < r < 1

1, if r = 1
This is called a GEOMETRIC SEQUENCE.
Example 5.11. Some quick examples
n
2
2
1. lim
= 0 since −1 < < 1
n→∞
3
3
5n
2. lim n = lim
n→∞
n→∞ 3
n
5
5
diverges because > 1
3
3
3. lim 1n = 1 since r = 1
n→∞
4. lim
n→∞
−7
8
n
− 5 = 0 − 5 converges to −5
Example 5.12. Does an =
n2 − 4n
converge?
7n
Let’s check its limit.
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5.1 Sequences
Brian E. Veitch
n2 − 4n
n→∞
7n
2n − 4n ln 4
n→∞
7n ln 7
2 − 4n (ln 4)2
LH
→ lim
n→∞
7n (ln 7)2
−4n (ln 4)3
LH
→ lim n
n→∞ 7 (ln 7)2
3 n
4
ln 4
·
= lim −
n→∞
ln 7
7
4
= 0 since − 1 < < 1
7
LH
→
lim
So yes, an =
lim
n2 − 4n
does converge.
7n
Example 5.13. Does an =
5n+4
converge?
3n−2
We need to rewrite this so it’s of the form rn .
5n+4
=
n→∞ 3n−2
54 · 5n
n→∞ 3−2 · 3n
n
5
4
2
= lim 5 · 3 ·
n→∞
3
5
= ∞ since > 1
3
lim
Example 5.14. Does an =
lim
cos2 (n)
converge?
2n
We know
−1 ≤ cos2 (n) ≤ 1
Divide all sides by 2n
−
Since lim −
n→∞
1
cos2 (x)
1
≤
≤ n
n
n
2
2
2
1
1
cos2 (n)
=
lim
=
0,
then
by
the
Squeeze
Theorem,
lim
=0
n→∞
2n n→∞ 2n
2n
.
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5.1 Sequences
Brian E. Veitch
One of the biggest skills you can develop during the sequence and series section is to be
able to look at a sequence and make an educated (and hopefully) correct guess on whether
the sequence converges.
Here’s a list of terms you’ll come across and how they rank among each other when n is
very large.
c < ln n < any polynomial < exponential functions (base > 1) < n! < nn
An example would be
5 < 5 ln n200 + 515n3 < 3n < n! < nn
Example 5.15. Does an =
n!
converge?
nn
Based on what I wrote above, this should converge to 0. But let’s go ahead and prove it.
Let’s take a look at some of the terms.
• a1 =
1
1
• a2 =
2·1
2·2
• a3 =
3·2·1
3·3·3
Following the pattern, we can write
an =
n(n − 1)(n − 2)(n − 3) · ... · 3 · 2 · 1
n · n · n · n · ... · n · n
Each term on the numerator pairs with an n on the denominator.
an =
n n−1 n−2
3 2 1
·
·
· ... · · ·
n
n
n
n n n
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Brian E. Veitch
Rewrite it as
an =
1 2 3
n−2 n−1 n
· · · ... ·
·
·
n n n
n
n
n
1
an =
n
2 3
n−2 n−1 n
· · ... ·
·
·
n n
n
n
n
Since every product in the parentheses is ≤ 1, if you get rid of them, we have
0 < an ≤
1
n
1
= 0, we have
n→∞ n
By the Squeeze Theorem, since lim
n!
=0
n→∞ nn
lim
We’re almost done. Let’s just go over a couple more definitions before we move on.
Definition 5.4. A sequence is increasing if an < an+1 for n ≥ 1.
A sequence is decreasing if an > an+1 for n ≥ 1.
A sequence is monotonic if it is always increasing or always decreasing.
Example 5.16. an =
3
.
n+5
If you write out the first few terms, you’ll see an is decreasing to 0.
3 3 3 3 3
, , , , , ...,
6 7 8 9 10
So an a decreasing monotonic sequence.
1
an = 1 −
n
The first few terms are
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5.1 Sequences
Brian E. Veitch
1 2 3 4 5 6
0, , , , , , , ...
2 3 4 5 6 7
So an converges to 1. Since it’s always increasing, it’s monotonic.
(−1)n
an =
n
The first few terms are
1 −1 1 −1 1 −1
−1, ,
, ,
, ,
, ...
2 3 4 5 6 7
We can see an converges to 0. But it’s not always increasing and not always decreasing.
This is an example of a non-monotonic sequence.
Definition 5.5. A sequence is bounded above if there is a number M such that an ≤ M for
n ≥ 1.
A sequence is bounded below if there is a number m such that an ≥ m for n ≥ 1.
A sequence is called bounded if it is bounded from above and below.
Example 5.17.
1. an = 1 −
2. an =
1
is bounded below by 0 and above by 1.
n
(−1)n
is bounded below by -1, and above 1.
n
3. an = n is bounde below by 1, but it is not bounded above since lim n = ∞
n→∞
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