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1
AP Calculus BC
Review: Sequences, Infinite Series, and Convergence
Sequences
A sequence {𝒂𝒏 } is a function whose domain is the set of positive integers.
The functional values a1, a2, a3, . . . an are called the terms of the sequence.
The number an is called the nth term of the sequence.
A sequence is either convergent (if the sequence has a limit, i.e. it approaches a specific number)
or divergent (if the sequence does not approach a specific limit)
A sequence {𝑎𝑛 } has the limit L
DEF 1
𝑙𝑖𝑚 𝑎𝑛 = 𝐿
𝑛→∞
𝑜𝑟
𝑎𝑛 → 𝐿
𝑎𝑠
𝑛 →∞
if we can make the terms 𝑎𝑛 as close to 𝐿 as we like by taking 𝑛 sufficiently large.
If 𝑙𝑖𝑚 𝑎𝑛 exists , we say the sequence is convergent. Otherwise, we say the sequence is divergent.
𝑛→∞
DEF 2
𝑙𝑖𝑚 𝑎𝑛 = ∞
𝑛→∞
means that for every positive number 𝑀 there is an integer N such that 𝑎𝑛 > 𝑀 whenever n > N
It is a special type of divergence. We say {𝑎𝑛 } diverges to ∞ .
● Sequence {−1, 1, −1, 1, −1, 1, −1, … } is divergent, but doesn’t diverges to ∞ .
● Sequence 𝑎𝑛 =
𝑛2 + 3
is divergent too, and it does diverges to ∞ .
2𝑛 − 5
If {𝑎𝑛 } and {𝑏𝑛 } are convergent sequences and c is a constant, then
THM 1
𝑙𝑖𝑚 (𝑎𝑛 ± 𝑏𝑛 ) = 𝑙𝑖𝑚 𝑎𝑛 ± 𝑙𝑖𝑚 𝑏𝑛
𝑛→∞
𝑛→∞
𝑛→∞
𝑙𝑖𝑚 𝑐𝑎𝑛 = 𝑐 𝑙𝑖𝑚 𝑎𝑛
𝑛→∞
𝑛→∞
𝑙𝑖𝑚 (𝑎𝑛 𝑏𝑛 ) = 𝑙𝑖𝑚 𝑎𝑛 ∙ 𝑙𝑖𝑚 𝑏𝑛
𝑛→∞
𝑙𝑖𝑚
𝑛→∞
𝑛→∞
𝑛→∞
𝑝
𝑙𝑖𝑚 (𝑎𝑛 )𝑝 = (𝑙𝑖𝑚 𝑎𝑛 )
𝑛→∞
𝑛→∞
𝑙𝑖𝑚 𝑎𝑛
𝑎𝑛
= 𝑛→∞
𝑏𝑛 𝑛
𝑙𝑖𝑚 𝑏𝑛
𝑛→∞
𝑙𝑖𝑚
𝑛 make the terms
1 𝑎𝑛 as close𝑛→∞
if we can
to 𝐿1 as we like by 1taking 𝑛 sufficiently large.
● 𝑙𝑖𝑚
= 𝑙𝑖𝑚
=
=
= 1
𝑛→∞ 𝑛 + 1
𝑛→∞ 1 + 1⁄
1 + 0
𝑙𝑖𝑚
1 + 𝑙𝑖𝑚 1⁄
𝑛
𝑛
If 𝑙𝑖𝑚 𝑎𝑛 exists , we say the
sequence
is convergent.
𝑛→∞
𝑛→∞ Otherwise, we say the sequence is divergent.
𝑛→∞
THM 2 - 𝐓𝐡𝐞 𝐒𝐪𝐮𝐞𝐞𝐳𝐞 𝐭𝐡𝐦
If 𝑎𝑛 ≤ 𝑏𝑛 ≤ 𝑐𝑛 for 𝑛 ≥ 𝑛0 and 𝑙𝑖𝑚 𝑎𝑛 = 𝑙𝑖𝑚 𝑐𝑛 = 𝐿 , then 𝑙𝑖𝑚 𝑏𝑛 = 𝐿
𝑛→∞
● 𝑙𝑖𝑚
𝑛!
=?
𝑛𝑛
𝑎𝑛 =
1 ∙ 2 ∙ 3 ∙∙∙ 𝑛
1 1 ∙ 2 ∙ 3 ∙∙∙ 𝑛
= (
)
𝑛 ∙ 𝑛 ∙ 𝑛 ∙∙∙ 𝑛
𝑛 𝑛 ∙ 𝑛 ∙ 𝑛 ∙∙∙ 𝑛
𝑛→∞
𝑙𝑖𝑚 0 = 0
𝑛→∞
𝑙𝑖𝑚
1
𝑛→∞ 𝑛
= 0 (𝑇𝐻𝑀 2) ⟹
𝑛→∞
𝑛→∞
1 ∙ 2 ∙ 3 ∙∙∙ 𝑛
≤1 ⟹
𝑛 ∙ 𝑛 ∙ 𝑛 ∙∙∙ 𝑛
𝑙𝑖𝑚
𝑛!
𝑛→∞ 𝑛𝑛
=0
0 < 𝑎𝑛 ≤
1
𝑛
2
If lim 𝑓(𝑥) = 𝐿 𝑎𝑛𝑑 𝑓(𝑛) = 𝑎𝑛 , when n is an integer, then 𝑙𝑖𝑚 𝑎𝑛 = 𝐿
THM 3
𝑥→∞
● 𝑙𝑖𝑚
𝑛→∞
𝑛→∞
𝑙𝑛 𝑛
=?
𝑛
(
∞
)
∞
l'Hospital's rule cannot be applied to sequences but to function of real variable.
Introducing 𝑓(𝑥) = 𝑙𝑛 𝑥⁄𝑥 we can use it now.
𝑙𝑖𝑚
𝑥→∞
1⁄
𝑙𝑛 𝑥
𝑥
= 𝑙𝑖𝑚
= 0
𝑛→∞ 1
𝑥
● 𝑙𝑖𝑚
𝑛→∞
3𝑛
=?
𝑒 5𝑛
(
(𝑇𝐻𝑀 3) ⟹
𝑙𝑖𝑚
𝑛→∞
𝑙𝑛 𝑛
=0 ⟹
𝑛
∞
)
∞
Sequence is obviously approaching 0 as n approaches infinity. Let’s apply THM 3 anyway.
𝑓(𝑥) =
3𝑥
𝑒 5𝑥
3𝑥
∞
= ( )=
5𝑥
𝑥→∞ 𝑒
∞
lim 𝑓(𝑥) = lim
𝑥→∞
● Show that the sequence {
lim 𝑓(𝑥) = lim
𝑥→∞
𝑥→∞
1+2𝑛3
𝑛3
lim
𝑥→∞
3
=0 ⟹
5𝑒 5𝑥
𝑙𝑖𝑚
𝑛→∞
3𝑛
=0
𝑒 5𝑛
} converges
1 + 2𝑥 3
∞
= ( )=
3
𝑥
∞
lim
𝑥→∞
1/𝑥 3 + 2
=2 ⟹
1
1 + 2𝑛3
= 2
𝑛→∞
𝑛3
lim
We will often have to find a formula for the general, or nth, term of a sequence. Look at the next example:
● If the first four terms of a sequence {an} are 1 ,
9 27 81
,
,
7 11 15
a) find a formula for the nth term of the sequence
b) determine whether the sequence converges or diverges
a) sequence is:
31 32 33 34
3
nth term is: 𝑎𝑛 =
𝑏) 𝑓(𝑥) =
THM 4
,
7
3𝑛
, 11 , 15 , ⋯ , =
31
32
33
4𝑛−1
3𝑥
4𝑥 − 1
3𝑥
∞
3 𝑥 𝑙𝑛 3
= ( ) = lim
= ∞
𝑥→∞
4𝑥 − 1
∞
4
lim
𝑥→∞
If lim |𝑎𝑛 | = 0 , then 𝑙𝑖𝑚 𝑎𝑛 = 0
𝑥→∞
● 𝑙𝑖𝑚
𝑛→∞
𝑙𝑖𝑚 |
𝑛→∞
𝑛→∞
(−1)𝑛
=?
𝑛
(−1)𝑛
1
| = 𝑙𝑖𝑚
=0
𝑛→∞ 𝑛
𝑛
(𝑇𝐻𝑀 4) ⟹
𝑙𝑖𝑚
𝑛→∞
(−1)𝑛
=0
𝑛
● For what values of 𝑟 is the sequence {𝑟 𝑛 } convergent?
𝑟=0
𝑟=1
0<𝑟<1
−1 < 𝑟 < 0
34
,
,
,
,⋯
4(1)−1 4(2)−1 4(3)−1 4(4)−1
𝑙𝑖𝑚 0𝑛 = 0
𝑛→∞
𝑙𝑖𝑚 1𝑛 = 1
𝑛→∞
𝑙𝑖𝑚 𝑟 𝑛 = 0
𝑛→∞
𝑙𝑖𝑚 |𝑟 𝑛 | = 0
𝑛→∞
(𝑇𝐻𝑀 4) ⟹
𝑙𝑖𝑚 𝑟 𝑛 = 0
𝑛→∞
{𝑟 𝑛 } converges for − 1 < 𝑟 < 1 and diverges for |𝑟| > 1
⟹ the sequence diverges.
3
DEF 3 – MONOTONIC SEQUENCE
is a sequence {𝑎𝑛 } such that either it is
increasing, that is 𝑎𝑛+1 > 𝑎𝑛 for every 𝑛 ≥ 1 , or decreasing, that is 𝑎𝑛+1 < 𝑎𝑛 for every 𝑛 ≥ 1 .
𝑛
is decreasing.
𝑛2 + 1
𝑛+1
𝑛
⟹
< 2
⟹ (𝑛 + 1)(𝑛2 + 1) < 𝑛[(𝑛 + 1)2 + 1]
(𝑛 + 1)2 + 1
𝑛 +1
● Show that sequence 𝑎𝑛 =
𝑎𝑛+1 < 𝑎𝑛
⟹ 𝑛3 + 𝑛2 + 𝑛 + 1 < 𝑛3 + 2𝑛2 + 2 𝑛 ⟹ 1 < 𝑛2 + 𝑛
For 𝑛 ≥ 1 inequality [1 < 𝑛2 + 𝑛] is true, so 𝑎𝑛 =
𝑛
𝑛2 +1
is decreasing.
A sequence {𝑎𝑛 } is bounded above if there is a number 𝑀 such that
DEF 4
𝑎𝑛 ≤ 𝑀
for every 𝑛 ≥ 1
A sequence {𝑎𝑛 } is bounded below if there is a number 𝑚 such that
𝑎𝑛 ≥ 𝑚
for every 𝑛 ≥ 1
A sequence {𝑎𝑛 } is bounded if it is bounded above and below.
THM 5 – MONOTONIC SEQUENCE THEOREM
● A bounded monotonic increasing sequence is convergent (it converges to its greatest lower bound).
● A bounded monotonic decreasing sequence is convergent (it converges to its greatest lower bound).
● A convergent sequence is bound.
● There are sequences which are convergent without being monotonic. Sequences:
1
1 1
2
3 4
{−1, , − , , … }
and
1
{ ,
1
1 1
, ,
2 22, 3 32
, … } both converge to 0.
Mathematical induction is a method of mathematical proof typically used to establish that a given statement P(n)
is true for all natural numbers
1. step: verify that P(n) is true for n = i (i is usually 1)
2. step (inductive step): assuming that there is a 𝑘 ≥ 1, for which P(k) is true, then
3. step: prove that, P(k+1) is true.
Since you have verified P(i), it follows from the inductive step that P(i + 1) is true, and hence, P(i + 2) is true, and
hence P(i + 3) is true, and so on. In this way the theorem has been proved.
The Fibonacci numbers are recursively defined as 𝑎1 = 1, 𝑎2 = 1, 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 , 𝑛 > 2
Show that the sequence of Fibonacci numbers {1, 1, 2, 3, 5, 8, 13, 21, ...} does not converge.
We will show by induction that the sequence of Fibonacci numbers is unbounded. If that is true, then the sequence
can not converge, because every convergent sequence must be bounded.
We will show that the n-th term of that sequence is greater or equal to n, at least for n > 4.
Property P(n): 𝑎𝑛 ≥ 𝑛 for all 𝑛 > 4
1. Check the lowest term: 𝑎5 = 𝑎4 + 𝑎3 = 5 ≥ 5 is true
2. Assume: 𝑎𝑛 ≥ 𝑛 for all 𝑛 > 4
3. Prove: 𝑎𝑛+1 ≥ 𝑛 + 1
𝑎𝑛+1 = 𝑎𝑛 + 𝑎𝑛−1 ≥ 𝑛 + 𝑎 ≥ 𝑛 + 1 ≥ 𝑛 ,
Hence, by induction the Fibonacci numbers are unbounded and the sequence can not converge.