Download Series and BMCT • Suppose that {a k=m is a sequence of real

Series and BMCT
• Suppose that {ak }∞
k=m is a sequence of real numbers, where m is fixed.
Then
∞
n
X
X
def
ak = lim
ak .
n→∞
k=m
k=m
If limn→∞ k=m ak = L exists, then we say the series ∞
k=m ak conP
verges with sum L. If the limit of nk=m ak does not exist, then we
P
say the series ∞
k=m ak diverges.
Pn
P
• The sum of the first n-terms in a series is called the n-th partial sum
P
of the series. For instance, nk=m ak is the (n − m + 1)-th partial sum
P∞
of the series k=m ak .
• To determine whether the partial sum sn =
may evaluate sn and try the following:
Pn
k=m ak
converges, one
– Check whether there is a number L such that for every ε > 0,
there exists a number N such that
n > N ⇒ |sn − L| < ε.
– Use limit rules to determine whether sn converges. (squeeze rule,
composition limit rule...)
– If sn = f (n) for some f (x), check whether limx→∞ f (x) exists.
• Examlpes of finding the sum of a series through direct caculation.
Example 1. Find
Sol. Recall that
∞
X
P∞
−k
k=1 2 .
Pn
k=1 ar
2−k = lim
k=1
n→∞
k
n
X
P∞
0.5(1 − (0.5)n )
= 1.
n→∞
1 − 0.5
2−k = lim
k=1
– The geometric series
Example 2. Find
= ar(1 − rn )/(1 − r) if r 6= 1. Therefore,
P∞
k=3 (2k
k=m ar
k
= arm /(1 − r) if |r| < 1.
+ 1)/(k 2 (k + 1)2 ).
• Divergence test (Theorem 8.9). If limk→∞ ak 6= 0, then
verges.
1
P∞
k=m ak
di-
Example 3. Suppose that a 6= 0. Then the geometric series
diverges if |r| ≥ 1.
P∞
k=m ar
k
• Another important result for determining the convergence of a sequence is the BMCT (bounded, monotonic, convergence theorem),
which states that a monotone sequence is convergent if and only if
it is bounded.
• Bounded sequences.
– A sequence {bn } is bounded above if there exists a number M
such that bn ≤ M for every n. In such case, M is called an upper
bound of {bn }.
– A sequence {bn } is bounded below if there exists a number M
such that bn ≥ M for every n. In such case, M is called a lower
bound of {bn }.
– A sequence {bn } is bounded if it is bounded both above and
below.
• Monotone sequences.
– A sequence {bn } is increasing if bn ≤ bn+1 for every n. If bn <
bn+1 for every n, then {bn } is strictly increasing.
– A sequence {bn } is decreasing if bn ≥ bn+1 for every n. If bn >
bn+1 for every n, then {bn } is strictly decreasing.
– A sequence is monotone (or monotonic) if the sequence is increasing or decreasing.
• The BMCT follows from the following fact.
Fact 1. Suppose that a sequence {bn } is increasing. Then
(i) {bn } converges to its least upper bound if {bn } is bounded above,
and
(ii) {bn } diverges to ∞ if it is not bounded above.
• Least upper bounds. Suppose that S is a non-empty subset of R =
(−∞, ∞).
– A number M is called an upper bound of S if for every x ∈ S,
x ≤ M . If S has an upper bound, then we say it is bounded
above.
2
– A number L is called the least upper bound of S if
(i) for every x ∈ S, x ≤ L, and
(ii) for every ε > 0, there exists some x ∈ S such that x > L − ε.
– If S is a non-empty subset of R and S is bounded above, then S
has a least upper bound.
Example 4. The least upper bound of the interval [0, 1] is 1.
Example 5. The least upper bound of the interval [0, 1) is 1.
Example 6.
The least upper bound of S = {1 − n−1 : n ∈
{1, 2, . . .}} is 1.
• BMCT/Fact 1 applications
Example 7. Show that the sequence {(1 + n−1 )n } is convergent.
Sol. Note that
(1 + (n + 1)−1 )n+1
(1 + n−1 )n
n
1
1
=
1−
1+
2
(n + 1)
n+1
n
1
≥
1−
1
+
(n + 1)2
n+1
3
2
n + 3n + 3n + 2
=
≥1
(n + 1)3
and
(1 + n−1 )n ≤
n
X
1
k!
k=0
≤
n
X
2
2k
k=0
< 4.
Thus {(1 + n−1 )n } is increasing and bounded above. By Fact 1, {(1 +
n−1 )n } is convergent.
√
√
Example 8. Suppose that a1 = 2 and an+1 = 2 + an for n ≥ 1.
Show that the sequence {an }∞
n=1 is convergent.
Example 9. Show that the series
P∞
k=0 1/k!
is convergent.
• Suppose that ak ≥ 0 for k ≥ m. Then the series ∞
if
k=m ak converges
P
P
the partial sum sequence { nk=m ak } is bounded above. If { nk=m ak }
P
is not bounded above, then ∞
k=m ak diverges to ∞.
P
3