1 Calc II-Dr. Almus
Math 2414 Calculus II
Dr. Melahat Almus
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Section 9.2 Convergence of a Sequence
Definition: Limit of a Sequence:
We say that lim nn
a L
if for every 0 , there exists a natural number N such
that nn N a L .
In this case, the sequence na is said to be “convergent”. Otherwise, the sequence
is said to be “divergent”.
OR:
L is the limit of the sequence na
if and only if
for every positive number epsilon, no matter how small,
there is a natural number N such that
all terms of the sequence after Na stay within epsilon units of L.
(That is, if the terms of the sequence approach a unique target number L, we say
the limit is L.)
Example: 1
nan
; 1
nlim
n
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Example: 11 n
n
n
lim
1n
n
n
Example: 2 1 4 9 16 25 36na n , , , , , ,.... ; nnlim a
Example: 1 11 11 11n
na , , , , , ,... ; nnlim a
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Let na be a sequence of real numbers. If the sequence has a finite limit, we say
it is convergent. If it does not have a finite limit, we say it is divergent.
Possibilities:
1) lim nn
a L
where L is a finite real number. We say the sequence na
converges to L.
2) lim nn
a
; we say the sequence diverges to infinity.
3) lim nn
a
; we say the sequence diverges to negative infinity.
4) lim :nn
a DNE
because the sequence oscillates between two or more
numbers. We say the sequence is divergent due to oscillation.
We worked on some simple examples and observed the convergence. What if the
general term of the sequence is more complicated? For example:
2
2
5
1n
na
n
We will work on more complicated cases in a bit; but first, some results:
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Theorem 9.2.4: Every convergent sequence is bounded.
That is:
If na is convergent, then it is bounded.
If na is bounded, then it may be convergent OR divergent.
IMPORTANT: If na is not bounded, then it is divergent.
(If a sequence is NOT bounded, you can use this fact to prove that it is divergent!)
Theorem 9.2.6:
A bounded, increasing sequence converges to its least upper bound.
A bounded, decreasing sequence converges to its greatest lower bound.
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Example: Determine whether the sequence is convergent or divergent. If
convergent, find the limit.
1
2n
n
2
11
nn
21
1
nn
1
1n
nn
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For rational expressions of the form polynomial
polynomial :
0n
nn
alim
b if the degree of denominator is bigger.
leading coefficient of numerator
leading coefficient of denom.
n
nn
alim
b if the degree of top and bottom are
equal.
n
n
a
b if the degree of top is bigger (sequence is not bounded, so divergent!).
2
1
2 1
n
n
n
3
2 3
1
4 1
5n
n
n n
3
1
1
n
n
n
10 8
12 8
1
5 2 1
4n
n n
n n n
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This “trick” works even if there are radicals; compare the largest exponents:
1
2 1
5 4n
n
n
3 2
3 2
1
5
2
/
/
n
n n
n
For more complicated rational formulas, you can compare how fast the top
and bottom are growing to decide whether the limit exists or not. If the
denominator grows faster, the limit is 0. If the numerator grows faster, the
sequence is divergent.
Fastest to slowest growing expressions: 2 3 >> ! >> exponentials (2 , ,..) >> polynomials( , , .)n n nn n e n n etc
2 3
Fastest:
Then: !
Then: exponentials (2 , ,..)
Then: polynomials( , , .)
n
n n
n
n
e
n n etc
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Examples: Determine if the sequence is convergent or divergent. If
convergent, find the limit.
1
2
! n
n
n
21
!
2 1 n
n
n
1
!
2nn
n
1
4
!
n
nn
1!
n
n
n
n
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Examples: Determine if the sequence is convergent or divergent. If
convergent, find the limit.
10
1
2n
n
n
e
1
!
2nn
n
n
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Remark: If each term (after the first term) of a sequence is obtained by
multiplying the previous term by a constant number (different from 1), then that
sequence is called a geometric sequence.
The sequences of the type: 2 3 4, , , ,....nar ar ar ar ar are geometric sequences
(where r is a real number different from 1.)
Example: 2 2,4,8,16,32,....n
1 1 1 1 1, , , ,....
10 10 100 1000 10000
n
FACT: The sequences of the form nr where r is a real number ( 1r ) are
geometric sequences and we know their limit:
For geometric sequences:
If 1r , then 0nr as n .
If 1r , then nr diverges.
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Example: Determine the limit of the sequences:
1
2
5
n
n
1
3
2
n
n
1
4n
n
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Example: Find the limit of the sequence: 32
5
n
n
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Note: The sequences of the type: 2 3 4, , , ,....nar ar ar ar ar are also geometric.
And their limit is still easy to find:
lim 0n
nar
if 1r .
Example: Find the limit of the sequence: 5 5 5 5
, , ,...2 4 82n
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Oscillating Sequences
A sequence is “oscillating” if its terms are accumulating around more than 1 finite
number. Oscillating sequences are DIVERGENT.
Example: 1
1 11 11 11n
n, , , , , ,.....
This sequence is oscillating because it accumulates around “1” and “-1”.
The following are common oscillating examples:
1
cos 1,1, 1,1, 1,1,...n
n
1n
sin n
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Study the following examples:
Example: The sequence 10 ( 1)
4
n n
n
If n is odd: the formula looks like: 10 1
4 4
n
n
If n is even: the formula looks like: 10 1
4 4
n
n
Hence, the sequence is oscillating and divergent.
Example: The sequence 2
10 ( 1)
4
n n
n
If n is odd: the formula looks like: 2
100
4
n
n
If n is even: the formula looks like: 2
100
4
n
n
Hence, the sequence is NOT oscillating. It converges to 0.
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Some Important Limits:
11
n
nlim e
n
and 1
bnab
n
alim e
n
1
1/ n
nlim n
By Theorem 9.2.12:
lim lim ln lnn nn n
a L a L
lim lim sin sinn nn n
a L a L
lim lim cos cosn nn n
a L a L
For each real 0x , lim 0!
n
n
x
n
. Example:
2lim 0
!
n
n n
For each real 0x , 1/lim 1n
nx
. Example: 1/lim 5 1n
n
0
n
sin nlim
n
0
n
ln nlim
n
The sequence nr converges to 0 if 1r , diverges if 1r .
Note: The sequences of the form nr (where r is a nonzero number) are called “geometric
sequences”.
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READ THIS A NOTE ON THE FACT: 1
1
n
nlim e
n
What’s the reason?
Recall from Calculus 1:
Example: 1
1
x
xlim ?
x
This is an indeterminate form of the type 1 .
We say: 1 1
1 ln( ) ln 1
x
y y xx x
2
2
1ln 1
1 0lim ln( ) lim ln 1 lim Use L'hospital's Rule
1 0
1
11
1= lim lim 1
1 11
x x x
x x
xy x
x
x
x
x
x x
Now, the original limit is (need to convert!):
11lim lim 1
x
x xy e e
x
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MISCALLENOUS EXAMPLES
Example:
52
lim 1 ?
n
n n
Example: Find the limit of the sequence if it converges.
42
1
n
nlim
n
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Example: Find the limit of the sequence if it converges.
2
n
n
nlim
n
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Example: Give the limit if exists:
1
2nn
Example: Give the limit if exists:
1
52n
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Example: Give the limit if exists:
23
4
n
n
Example: Find the limit if it exists: 1
3
1
5
2
n
n
n
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Example: Give the limit if exists: 2ln 5 1n
Example: Give the limit if exists: ln
2
n
n
Example: Give the limit if exists: 2 2ln 4 1 ln 2n n
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Example: Give the limit if exists: sin
5
n
n
Example: Give the limit if exists: 2
sinn
Example: Give the limit if exists: 2
sin n
Example: Give the limit if exists: 4 5
ntan
n
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Exercise: Find the limit of the sequences with general term: 4
0
n
xna e
Exercise: Find the limit of the sequences with general term:
2
1
1
n
n
n
a dxx
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