(Comparison Theorem).

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Oct 8, 2013 (3 years and 10 months ago)

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ASYMPTOTICS AND COMPARISON THEOREMS
TOM BELULOVICH
The Comparison Theorem
Theorem 1 (Comparison Theorem).Let 0  a
n
 b
n
be sequences.
Then if
P
1
n=1
b
n
converges,the series
P
1
n=1
a
n
also converges.Likewise,
if
P
1
n=1
a
n
diverges,so does
P
1
n=1
b
n
.
Proof.Sums respect inequalities.The only way a series of nonnegative
terms can diverge is for it to diverge to innity.If
P
1
n=1
b
n
is nite,it
is certainly true that
P
1
n=1
a
n

P
1
n=1
b
n
< 1is also nite.Therefore
P
1
n=1
a
n
converges.
Unfortunately,direct application of this theoremcan be a bit clunky.
If we want to analyze some series
P
a
n
,we must rst guess whether
or not the series converges.If we want to prove convergence,we must
bound a
n
above by some sequence b
n
term-by-term,and show
P
b
n
<
1.If we want to prove divergence,we must bound a
n
below by some
sequence c
n
and show
P
c
n
= 1.
There are two issues with this approach:rstly,we'd have to guess
whether or not
P
a
n
converges.Secondly,we don't necessarily know
if the sequence we are comparing to is tightly-coupled enough to a
n
to be meaningful:for example,if we wanted to show
P
1
n=1
log(n) di-
verges,we could compare it to
P
1
n=1
1
2
n
,but this is a convergent sum,
which tells us nothing!It would be good to know the sequence we're
comparing to is the\right one."
Asecond issue that can pop up is that we might have a very appealing
sequence b
n
of terms to compare a
n
to,but perhaps a
n
 b
n
only
for n > N for some constant N.Certainly this doesn't aect the
convergence or divergence of
P
a
n
and
P
b
n
,but it isn't quite proper
to say
P
1
n=1
a
n

P
1
n=1
b
n
unless for every n we have a
n
 b
n
.
A solution:Asymptotics
We want to capture the idea that a
n
 b
n
eventually.We also would
like a guarantee that if
P
b
n
converges (diverges),that
P
a
n
converges
too (resp.diverges.) There turns out to be a way to compare the
1
2 TOM BELULOVICH
limiting behavior of sequences in such a way that it tells us something
about the convergence (or divergence) of the corresponding series.
Throughout,let a
n
and b
n
be sequences of real numbers that are
eventually nonnegative.This means there is some M > 0 where,for
every n > M,we have both a
n
 0 and b
n
 0.
Denition.The sequence a
n
weakly dominates b
n
,written a
n
& b
n
,if
there is an integer N and a real constant c > 0 such that,for every
n > N,we have
0  b
n
 ca
n
:
For example,n
2
 1 & n,since eventually n
2
 1 becomes much
greater than n,even though at n = 1 it is the case that n
2
1 < n.
Likewise,we want to be able to say n
2
& 2n
2
for reasons that will
become apparent later.
Denition.a
n
is weakly dominated by b
n
,written a
n
.b
n
,if b
n
& a
n
.
Denition.We say that a
n
is similar to b
n
,written a
n
 b
n
,if a
n
.b
n
and a
n
& b
n
.This means there are positive constants c and C and an
integer N such that,for n > N,we have
cb
n
 a
n
 Cb
n
:
There are strict versions as well:
Denition.The sequence a
n
dominates b
n
,written a
n
b
n
,if lim
n!1
b
n
a
n
=
0.
Denition.The sequence a
n
is dominated by b
n
,written a
n
b
n
,if
lim
n!1
b
n
a
n
= 1.Equivalently,b
n
a
n
.
Interpretation as Limits.There is a compact denition of the above
when the limit lim
n!1
a
n
b
n
exists,which it often will:
Proposition.Suppose L = lim
n!1
a
n
b
n
.Then
(1) If L > 0,then a
n
& b
n
.
(2) If L < 1,then a
n
.b
n
.
(3) If L = 0,then a
n
b
n
.
(4) If L = 1,then a
n
b
n
.
(5) If 0 < L < 1,then a
n
 b
n
.
Asymptotic Comparison Theorems and the Limit
Comparison Theorem
Theorem 2.Suppose a
n
.b
n
.Then if
P
1
n=1
b
n
converges,so does
P
1
n=1
a
n
.If
P
1
n=1
a
n
diverges,so does
P
1
n=1
b
n
.
ASYMPTOTICS AND COMPARISON THEOREMS 3
Theorem 3 (Limit Comparison Theorem).Suppose a
n
 b
n
.Then
P
1
n=1
a
n
converges if and only if
P
1
n=1
b
n
converges.
Computational Tools
We can add and multiply on both sides of.just like .
Proposition.
(1) If a
n
.b
n
and c
n
.d
n
,then a
n
+c
n
.b
n
+d
n
.
(2) If a
n
.b
n
and c
n
.d
n
,then a
n
c
n
.b
n
d
n
.
Proof.Let us prove part (1).There are positive constants c;d;N
1
;N
2
,
such that for n > N
1
,
0  a
n
 cb
n
and for n > N
2
,
0  c
n
 db
n
:
Then for n > max(N
1
;N
2
),it is the case that
0  a
n
+c
n
 cb
n
+db
n
 max(c;d)(b
n
+d
n
):
Therefore a
n
+c
n
.b
n
+d
n
.
We can even divide:
Proposition.If a
n
.b
n
,and a
n
;b
n
are eventually positive,then
1
a
n
&
1
b
n
.
The above two propositions are true for ,too!
Dominant Term Calculation.If f & g,then f is the dominant
term of f +g.
Proposition.If f & g,then (f +g)  f.
Proof.We have
f.(f +g).(f +f) = 2f.f
so f  (f +g).
Similarly,if we have m functions f
1
;:::;f
m
satisfying f
1
& f
i
for
every i,then
f
1
+f
2
+   +f
m
 f
1
:
This means that a nite sum of terms is always similar to its dominant
term.This is particularly useful.For example:
Proposition.Let p(n) = a
d
n
d
+    + a
0
be a degree d polynomial,
where a
d
6= 0.Then p(n)  n
d
.
4 TOM BELULOVICH
Fundamental Examples
Proposition.For any real constant q > 0,
ln(n) n
q
:
Proof.Apply limits.
lim
n!1
ln(n)
n
q
is an
1
1
indeterminate form,and so can be evaluated using L'Hopital's
rule.Doing so,we get
lim
n!1
1
n
qn
q1
= lim
n!1
1
qn
q
= 0
since q > 0.
In words,a logarithmic function is dominated by any positive power
of n.
Proposition.For any real constant q > 0,
n
q
e
n
:
Proof.Look at the limit
lim
n!1
n
q
e
n
= lim
n!1
qn
q1
e
n
by L'Hopital's rule.If q < 1,then this latter limit is zero.Otherwise,
the limit vanishes by induction on the hypothesis that n
q1
e
n
.
Examples
Example 1.Determine the convergence of
1
X
k=1
k
2
+3
k
3
1
:
Solution:Since k
2
& 3,we know k
2
+ 3  k
2
.Furthermore,
k
3
1  k
3
.Therefore this series converges if and only if
1
X
k=1
k
2
k
3
=
1
X
k=1
1
k
converges.But this is a
divergent
series by the p-test (specically,it
is the harmonic series.) 
ASYMPTOTICS AND COMPARISON THEOREMS 5
Example 2.Determine the convergence of
1
X
n=7
(n
2
+n +cos(n) +log(n))e
n
3
:
Solution:This looks like a mess!But n
2
 n
2
+n+cos(n) +log(n)
since n
2
dominates n;j cos(n)j;and log(n).Therefore n
2
e
n
3
 (n
2
+
n +cos(n) +log(n))e
n
3
,and it suces to check the convergence of
X
n=7
n
2
e
n
3
:
But
R
1
7
x
2
e
x
3
dx =
1
3
e
x
3



1
7
< 1,so by the integral test this series
is
convergent.