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Limit of a Sequence: − N Def.
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Definition.1 Let {an }∞
n=1 be a sequence.
(1) The limit of the sequence {an }∞
n=1 exists and is equal to the real number L, written
lim an = L
n→∞
or
n→∞
an −−−−→ L
provided:
for all positive numbers , there exists a natural number N , such that
if n is a natural number bigger than N , then L − < an < L + .
(1)
Since L − < an < L + if and only if |an − L| < , condition (1) is the same as
for all positive numbers , there exists a natural number N ,
if n > N , then |an − L| < .
such that
(2)
Condition (1) is also the same as
for all positive numbers , there exists a natural number N ,
for all n > N , |an − L| < .
such that
(3)
You are welcome to use that fun/ny shorthand notation from class, which was
∀ is shorthand for for all
∃ is shorthand for there exists,
to condensely rewrite the above conditions as:
∀ > 0, ∃ N ∈ N, such that
n>N
=⇒ L − < an < L + .
(10 )
∀ > 0, ∃ N ∈ N, such that
n>N
=⇒ |an − L| < .
(20 )
∀ > 0, ∃ N ∈ N, such that
∀n > N , |an − L| < .
(30 )
(2) {an }∞
n=1 converges (or is convergent) ⇔ you can find a real number L so that limn→∞ an = L.
(3) {an }∞
n=1 diverges (or is divergent)
⇔ {an }∞
n=1 does not converge.
1
For the definition of limn→∞ an = L, understand the picture/concept from class (that -band and then go pick
your N picture). You can express the definition using whichever of the 6 (equivalent) boxes that you prefer.
Prof. Girardi, Spring 2016
Page 1 of 2
Sequences
Limit of a Sequence: − N Def.
Example. Using the definition of limit of a sequence, show that
1
lim 4 −
= 4.
n→∞
17n + ln n
Solution.
Fix an arbitratry > 0.
Pick (and fix) N ∈ N so big that
N>
1
.
17
(∗)
Fix n ∈ N with n > N . Then
−1
1
1
1
=
4 −
−
4
=
≤
x
x
x
 17n + ln n
 17n + ln n
17n + ln n
 17n



<
x
1
17N






by
algebra
by
algebra
since
17n ≤ 17n+ln n
since
n>N
<
x .



by (∗),
i.e. by our
choice of N
We have just shown that for an arbitrary > 0, we can pick N ∈ N such that if n > N then
1
1
4 −
− 4 < . So, by definition of limit of a sequence, limn→∞ 4 − 17n+ln
= 4.
17n+ln n
n
Problem. Using the definition of limit of a sequence, show that
3n2 − sin n
3
lim
=
.
2
n→∞
16n
16
Prof. Girardi, Spring 2016
Page 2 of 2
Sequences