4.2. SERIES CONVERGENCE TESTS 131
Solution:
is series has the form of a p-series with p D 1=2. Since 1=2 1, we conclude the se-
ries diverges.
˙
e next two tests leverage series we already understand to resolve even more series. e tests
depend on various forms of comparison of a series under test to a series with known convergence
or divergence behavior.
Knowledge Box 4.11
e comparison tests
• If
1
X
nD0
x
n
converges and 0 y
n
x
n
for all n, we have that
1
X
nD0
y
n
also converges.
• If
1
X
nD0
x
n
diverges and y
n
x
n
0 for all n, we have that
1
X
nD0
y
n
also diverges.
e colloquial versions of the comparison tests are pretty easy to believe. e first says, of a series
with positive general terms, that if it is smaller, term by term, than another series with a finite
sum, then it has a finite sum. e second reverses that: saying that, if a series is larger, term by
term, than another series with an infinite sum, then it has an infinite sum.
Example 4.27 Determine if
1
X
nD1
1
n=3
converges or diverges.
132 4. SEQUENCES, SERIES, AND FUNCTION APPROXIMATION
Solution:
Notice that, for n 1,
n n=3
1
n
1
n=3
So 0
1
n
1
n=3
, which permits us to deduce that the series in the example diverges by
comparison to the harmonic series, which is known to diverge.
˙
Example 4.28 Determine if
1
X
nD0
1
2
n
C 5
converges or diverges.
Solution:
e key to using the comparison test is to find a good known series to compare to. In
this case the geometric series with ratio 1/2 is natural.
2
n
2
n
C 5
1
2
n
1
2
n
C 5
1
2
n
1
2
n
C 5
Which means that the general term of the convergent geometric series
1
X
nD0
1
2
n
is greater than the general term of our target series. Since all terms are positive, this tells us that
we may conclude the target series converges, by comparison.
˙
4.2. SERIES CONVERGENCE TESTS 133
e next test is one of the most all-around useful tests. It permits us to resolve any series drawn
from a rational function, for example, by checking to see which p-series a rational function is
most like.
Knowledge Box 4.12
e limit comparison test
Suppose that
1
X
nD0
x
n
and
1
X
nD0
y
n
are series, and that
lim
n!1
x
n
y
n
D C
where 0 < jC j < 1 is a constant. en, either both series converge or both series di-
verge.
Example 4.29 Determine if
1
X
nD1
n C5
n
3
C 1
converges or diverges.
Solution:
Perform limit comparison to
1
X
nD1
1
n
2
lim
x
n
y
n
D lim
n!1
nC5
n
3
C1
1
n
2
D lim
n!1
n C5
n
3
C 1
n
2
1
D lim
n!1
n
3
C 5n
2
n
3
C 1
D 1
Since 0 < 1 < 1, we can conclude that
1
X
nD1
n C5
n
3
C 1
converges, because
1
X
nD1
1
n
2
is a convergent
p-series.
˙
134 4. SEQUENCES, SERIES, AND FUNCTION APPROXIMATION
Notice that choosing a p-series with p D 2 was exactly what was needed to make the ratio
of the general terms of the series be something that had a finite, non-zero limit. e limit
comparison test is useful for putting rigour into the intuition that one series is “like” another.
e next test uses the fact that if you have a sum of positive terms that is finite, making some or
all of those terms negative cannot take the sum farther from zero than the original finite sum –
leaving the sum finite.
Knowledge Box 4.13
e absolute convergence test
If
1
X
nD0
jx
n
j converges, then so does
1
X
nD0
x
n
.
Example 4.30 Prove that
1
1
4
C
1
9
1
16
C
1
25
converges.
Solution:
e series in question is given as an obvious pattern – begin by pulling it into summa-
tion notation:
1
X
nD1
.1/
nC1
n
2
If we take the absolute value of the general terms of this series, we obtain the series:
1
X
nD1
1
n
2
We know this to be a convergent p-series. We may deduce that the series converges.
˙
4.2. SERIES CONVERGENCE TESTS 135
Definition 4.9 If
1
X
nD0
jx
n
j converges, then we say the series
1
X
nD0
x
n
converges in absolute value.
is means we can restate Knowledge Box 4.13 as: “A series that converges in absolute value,
converges.”
e next test uses the fact that if you go up, then down by less, then up by even less, and so on,
you end up a finite distance from your starting point.
Knowledge Box 4.14
e alternating series test
Suppose that x
n
is a series such that x
n
> x
nC1
0, and suppose that lim
n!1
x
n
D 0.
en the series
1
X
nD0
.1/
n
x
n
converges.
Example 4.31 Show that
1
X
nD1
.1/
n
p
n
converges.
Solution:
We already know that lim
n!1
1
p
n
D 0, and the terms of the series clearly get smaller as n
increases. So, we may conclude this series converges by the alternating series test.
˙
e next two tests both check to see if a series is like a geometric series and deduce its
convergence or divergence from that similarity.
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