comparison test
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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
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Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchanka
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.
EXAMPLE: Determine whether the series
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.
EXAMPLE: Determine whether the seriesn=2
1
n ln2 nconverges or diverges.
2
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Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchanka
EXAMPLE: Determine whether the series
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.
EXAMPLE: Determine whether the seriesn=1
1
n + 1converges or diverges.
3
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Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchanka
EXAMPLE: Determine whether the series
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
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Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchanka
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
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