Download Series, Part 1 - UCSD Mathematics

SERIES, PART 1
The idea of a series is to make sense of summing an in…nite sequence of numbers. Let fan g be
a sequence. For each n 2 N we can sum the …rst n terms of this sequence to get
sn =
n
X
ak = a1 + a2 +
+ an :
k=1
Then fsn g is also a sequence. sn is called the n-th partial sum.
P1
De…nition. If fsn g converges
to
a
number
s,
then
we
say
that
the
in…nite
series
P
P1 n=1 an converges to s and we write 1
a
=
s.
If
fs
g
diverges,
then
we
say
that
the
series
n
n=1 n
n=1 an diverges.
n
Example. Consider fan g, where an = 2 . Then
sn =
n
X
2
k
=
k=1
1 1
+ +
2 4
+
1
=1
2n
1
:
2n
P
k
= 1.
Since limn!1 sn = limn!1 1 21n = 1, we write 1
k=1 2
Pn
Lemma 1. Let fan g be a sequence andP
let sn = k=1 ak be the n-th partial sum. Suppose that
sn bn , where bn ! +1. Then the series 1
n=1 an diverges.
P
Proof. Let x 2 R be any real number. We prove that 1
n=1 an cannot converge to x. Since
bn ! +1, there exists N 2 N such that bn x + 1 for all n N . Since sn bn , this implies
x + 1 for all n
sn
N:
Clearly sn cannot
P converge to x. Since fsn g cannot converge to any real number x, it diverges. That
is, the series 1
n=1 an diverges.
P
1
Example. The harmonic series is the series 1
n=1 n . That is, we start with the sequence fan g,
where an = n1 , and we de…ne the n-th partial sums
sn =
n
X
1
k=1
k
=
1 1
+ +
1 2
+
1
:
n
P1 1
k
As an exercise, prove that s2k
+
1.
By
Lemma
1,
this
implies
that
the
series
n=1 n diverges.
2
For example, see Proof 1 in:
P
http://scipp.ucsc.edu/~haber/archives/physics116A10/harmapa.pdfthe series 1
n=1 an diverges.
Recall from Theorem 3.11 that a sequence of real numbers converges if and only if it is a Cauchy
sequence. Given a sequence fan g, we apply
P this theorem to the corresponding sequence of partial
sums fsn g to get the following. The series 1
n=1 an converges if and only if fsn g is a Cauchy sequence,
that is, if and only if:
For every " > 0 there exists an N 2 N such that if m
n
N , then jsm sn 1 j < " (we use
n 1 instead of n to make the following display look nicer). Now
sm
sn
1
=
m
X
ak
k=1
n 1
X
k=1
So we have the following.
1
ak =
m
X
k=n
ak :
Theorem 3.22.
if m n N , then
P1
n=1
an converges if and only if for every " > 0 there exists an N 2 N such that
m
X
ak < ":
k=n
As a special case of the theorem we may take m = n in the statement to get:
For every " > 0 there exists an N 2 N such that if n N , then jan j < ". That is, an ! 0. So we
have:
P1
Theorem 3.23. If
n=1 an converges, then an ! 0.
However, the converse is not true as can be seen by the harmonic series.
Generalizing a previous example, we have geometric series:
Theorem 3.26. If jxj < 1, then
1
X
1
:
xn =
1
x
n=0
If jxj 1, then the series diverges.
Proof. Recall that
1 + x + x2 +
Thus, since x 6= 1,
sn =
n
X
+ xn (1
xk = 1 + x + x 2 +
xn+1 :
x) = 1
+ xn =
k=0
1
xn+1
:
1 x
(1) If jxj < 1, then xn+1 ! 0, which in turn implies that sn ! 1 1 x .
(2) If jxj > 1, then xn+1 diverges, so that fsn g diverges.
(3) If x = 1, then sn = n + 1, which diverges.
(4) If x = 1, then sn = 1 if n is even and sn = 0 if n is odd, which diverges.
Power functions often come up in series. We have:
Theorem 3.28. Let p be a real number.
P
1
(1) If p > 1, then 1
n=1 p converges.
n
P1 1
(2) If p 1, then n=1 p diverges.
n
Proof. See p. 62 of Rudin. (In calculus, you may have seen a ‘proof’using the integral test.)
Let fbn g be a sequence. Recall that lim supn!1 bn and lim inf n!1 bn are the least upper bound
and greatest lower bound of all subsequential limits, respectively.
Lemma 2 (a.k.a. Theorem 3.17(b)). Let = lim supn!1 bn and assume that 2 R. Then,
for each > 0 there exists N 2 N such that
bn <
for n
+
N:
Proof. Suppose the lemma is false. Then there exists > 0 with the property that there does not
exist N 2 N such that bn < + for all n N .
Claim. There exists a subsequence fbnk g1
+ for all k 2 N.
k=1 such that bnk
2
Proof of the claim. Firstly, by the property of for N = 1, we have bn1
+ for some n1 1.
Secondly, by the property of for N = n1 + 1, we have bn2
+ for some n2 n1 + 1. The claim
easily follows from mathematical induction.
With the claim we can obtain a contradiction as follows.
Case 1. fbnk g1
+ for all k 2 N, this implies there exists a
k=1 is unbounded. Since bnk
subsequence diverging to +1. This implies = +1, contradicting the assumption that 2 R.
Case 2. fbnk g1
k=1 is bounded. Then, by Theorem 3.6(b), there exists a convergent subsequence.
Since bnk
+ for all k 2 N, the subsequence converges to some number
+ . Since
,
this implies
+ , which is a contradiction.
In either case, we obtain a contradiction to the assumption that the lemma is false.
The idea of (the proof of) the ratio test is comparison with a geometric series.
P
Theorem 3.34 (Ratio Test) A series 1
n=1 an with an 6= 0 for all n
an+1
(a) converges if lim supn!1
< 1,
an
an+1
1 for all n n0 , where n0 2 N. That is, fjan jg is nondecreasing.
(b) diverges if
an
an+1
Proof. Part (a). Let = lim supn!1
and let = +1
. Since < 1, we have < < 1.
2
an
By Lemma 2, there exists N 2 N such that
an+1
<
an
for n
N:
That is,
jan+1 j <
jan j
for n
N:
By induction, we can prove that
k
jaN j
for k
n N
jaN j
for n > N:
jaN +k j <
1:
In other words,
Now the series
P1
jan j <
n N
P1
jaN j
n
converges since it is a geometric series (N is …xed)
P
and since 0 < < 1. Therefore, by (1) and the comparison test, 1
n=1 an converges.
an+1
1 for all n n0 , we have
Part (b). Since
an
n=N
jaN j =
(1)
n=N
jan+1 j
N
jan j
for n
n0 :
jan0 j
for n
n0 :
By induction, we can prove that
jan j
Since jan0 j > 0, we cannot have an ! 0. By Theorem 3.23 (its contrapositive), we conclude that
P
1
n=1 an diverges.
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