comparison test

Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchanka

The Integral and Comparison Tests

The Integral Test

THE INTEGRAL TEST: Suppose f is a continuous, positive, decreasing function on [1,)and let an = f(n). Then the series

n=1

an is convergent if and only if the improper integral

1

f(x)dx is convergent. In other words:

(a) If

1

f(x)dx is convergent, thenn=1

an is convergent.

(b) If

1

f(x)dx is divergent, then

n=1

an is divergent.

REMARK: Dont use the Integral Test to evaluate series, because in general

n=1

an 6=

1

f(x)dx

EXAMPLES:

1.

n=1

1

nis divergent, because f(x) =

1

xis continuous, positive, decreasing and

1

1

xdx is

divergent by the p-test for improper integrals, since p = 1 1.

2.n=1

1

n1/2is divergent, because f(x) =

1

x1/2is continuous, positive, decreasing and

1

1

x1/2dx

is divergent by the p-test for improper integrals, since p = 1/2 1.

3.n=1

1

n2is convergent, because f(x) =

1

x2is continuous, positive, decreasing and

1

1

x2dx is

convergent by the p-test for improper integrals, since p = 2 > 1.

REMARK 1: When we use the Integral Test it is not necessary to start the series or the integralat n = 1. For instance, in testing the series

n=5

1

n + 1we use

5

1

x+ 1dx

REMARK 2: It is not necessary that f be always decreasing. It has to be ultimately decreasing,that is, decreasing for x larger than some number N.

EXAMPLE: Determine whether the seriesn=1

lnn

nconverges or diverges.

1


EXAMPLE: Determine whether the series

n=1

lnn

nconverges or diverges.

Solution: The function f(x) =ln x

xis positive and continuous for x > 1. However, looking at

the graph of this function we conclude that f is not decreasing.

At the same time one can show that this function is ultimately decreasing. In fact,

f (x) =

(ln x

x

)=

(ln x) x lnx xx2

=1x x ln x 1

x2=

1 ln xx2

Note that 1 ln x < 0 for all sufficiently large x which means that f (x) < 0 and therefore fis ultimately decreasing. So, we can apply the Integral Test:

1

ln x

xdx = lim

t

t1

ln x

xdx = lim

t

(lnx)2

2

]t1

= limt

(ln t)2

2=

Since this integral diverges, the series

n=1

lnn

nalso diverges.


n=2

1

n lnnconverges or diverges.

Solution: The function f(x) =1

x ln xis continuous, positive and decreasing on [2,), therefore

we can apply the Integral Test:

2

1

x ln xdx = lim

t

t2

1

x ln xdx = lim

tln(ln x)]t2 = limt

[ln(ln t) ln(ln 2)] =

Since this integral diverges, the seriesn=2

1

n lnnalso diverges.


1

n ln2 nconverges or diverges.

2



n=2

1

n ln2 nconverges or diverges.

Solution: The function f(x) =1

x ln2 xis continuous, positive and decreasing on [2,), therefore

we can apply the Integral Test:

2

1

x ln2 xdx = lim

t

t2

1

x ln2 xdx = lim

t

[ 1ln x

]t2

= limt

[ 1ln t

+1

ln 2

]=

1

ln 2

Since this integral converges, the series

n=2

1

n ln2 nalso converges. We also note that, as it was

mentioned before,

n=2

1

n ln2 n6= 1

ln 2.

THEOREM (p-Test): The p-series

n=1

1

npis convergent if p > 1 and divergent if p 1.

Proof: We distinguish three cases:

Case I: If p < 0, then limn

1

np=, therefore

n=1

1

npdiverges by the Divergence Test.

Case II: If p = 0, then limn

1

np= lim

n

1

n0= lim

n

1

1= 1, therefore

n=1

1

npdiverges by the

Divergence Test.

Case III: If p > 0, then the function f(x) =1

xpis continuous, positive and decreasing on

[1,), therefore we can apply the Integral Test by whichn=1

1

npis convergent if and only if

the improper integral

1

1

xpdx is convergent. But

1

1

xpdx is convergent if p > 1 and divergent

if p 1 by the p-test for improper integrals.

REMARK: As before, when we use the p-Test it is not necessary to start the series at n = 1.

EXAMPLE: Determine whether the following series converge or diverge:

(a)n=1

1

n(b)

n=1

1

n2(c)

n=1

1

n1+, > 0

Solution: The series

n=1

1

ndiverges by the p-test for series, since p = 1 1. The series

n=1

1

n2

converges by the p-test for series, since p = 2 > 1. The seriesn=1

1

n1+converges by the p-test

for series, since p = 1 + > 1.


1

n + 1converges or diverges.

3



n=1

1

n + 1converges or diverges.

REMARK: Note that we CANT apply the p-test directly!

Solution 1: The function f(x) =1

x+ 1is continuous, positive and decreasing on [1,), there-

fore we can apply the Integral Test:

1

1

x+ 1dx = lim

t

t1

1

x+ 1dx = lim

tln(x+ 1)]t1 = limt

[ln(t+ 1) ln 2] =

Since this integral diverges, the seriesn=1

1

n+ 1also diverges.

Solution 2: We have n=1

1

n + 1=

n=2

1

n

Since

n=2

1

ndiverges by the p-test with p = 1, it follows that

n=1

1

n+ 1also diverges, since

convergence or divergence is unaffected by deleting a finite number of terms.

The Comparison Tests

THE COMPARISON TEST: Suppose that

an and

bn are series with positive terms.

(a) If

bn is convergent and an bn for all n, then

an is also convergent

(b) If

bn is divergent and an bn for all n, then

an is also divergent

EXAMPLE: Use the Comparison Test to determine whether the following series converge ordiverge.

(a)

n=1

lnn

n(b)

n=2

13n 1

Solution: Sincelnn

n>

1

nfor n > e and

n=1

1

ndiverges by the p-test with p = 1, it follows that

n=1

lnn

nalso diverges. Similarly, since

13n 1 >

13nand

n=2

13ndiverges by the p-test with

p = 1/3, it follows that

n=2

13n 1 also diverges.

EXAMPLE: Use the Comparison Test to determine whether the following series converge ordiverge:

(a)

n=1

1

n3 + n + 1(b)

n=2

1

n2 + 1(c)

n=2

1

n2 1

4


EXAMPLE: Use the Comparison Test to determine whether the following series converge ordiverge:

(a)

n=1

1

n3 + n + 1(b)

n=2

1

n2 + 1(c)

n=2

1

n2 1

Solution:

(a) Since

1

n3 + n+ 1

comparison test

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