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CM221A
ANALYSIS I
NOTES ON WEEK 4
TWO CONVERGENCE TESTS
Ratio Test Theorem.
Assume that
a
n
6
= 0 for all
n
, and that lim
n
→∞

a
n
+1
/a
n

exists and is equal to
c
. If
c <
1 then the series
∑
∞
n
=1
a
n
is absolutely convergent.
If
c >
1 then the series diverges. (If
c
= 1 nothing can be said.)
Proof.
Assume that
c <
1. Then there exists a positive number
b
such that
c < b <
1. From the deﬁnition of the limit it follows that

a
n
+1
/a
n

6
b
for all
suﬃciently large
n
or, in other words,

a
n
+1

6
b

a
n

for all
n
>
n
0
, where
n
0
is
some positive integer depending on
b
. Then we have

a
n
0
+1

6
b

a
n
0

,

a
n
0
+2

6
b
2

a
n
0

,

a
n
0
+3

6
b
3

a
n
0

, ...,

a
n
0
+
j

6
b
j

a
n
0

, ...,
which implies that

a
n

6

a
n
0

b

n
0
b
n
for all
n
>
n
0
. Since the series
∑
∞
n
=1
b
n
converges, by the Comparison Theorem the series
∑
∞
n
=1
a
n
is absolutely convergent.
Assume now that
c >
1. Then there exists a positive number
b
such that 1
< b < c
.
From the deﬁnition of the limit it follows that

a
n
+1
/a
n

>
b
for all suﬃciently large
n
or, in other words,

a
n
+1

>
b

a
n

for all
n
>
n
0
, where
n
0
is some positive integer
depending on
b
. Then we have

a
n
0
+1

>
b

a
n
0

,

a
n
0
+2

>
b
2

a
n
0

,

a
n
0
+3

>
b
3

a
n
0

, ...,

a
n
0
+
j

>
b
j

a
n
0

, ...,
which implies that
a
n
0
+
k
→
+
∞
as
k
→ ∞
. In this case the series
∑
∞
n
=1
a
n
diverges
because
a
n
do not converge to 0 as
n
→ ∞
.
n
throot Test Theorem.
Let lim sup

a
n

1
/n
=
c
. If
c <
1 then the series
∑
∞
n
=1
a
n
is absolutely convergent. If
c >
1 then the series diverges. (If
c
= 1 nothing can
be said.)
Proof.
Assume that
c <
1. Then there exists a positive number
b
such that
c < b <
1. From the deﬁnition of the upper limit it follows that
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 Spring '09

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