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 innity.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 aect 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.

Denition.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.

Denition.a

n

is weakly dominated by b

n

,written a

n

.b

n

,if b

n

& a

n

.

Denition.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:

Denition.The sequence a

n

dominates b

n

,written a

n

b

n

,if lim

n!1

b

n

a

n

=

0.

Denition.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 denition 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

q1

= 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

q1

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

q1

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 (specically,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 suces 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.

## Comments 0

Log in to post a comment