Exercise 2.5.7
Let
{
a
n
}
,
{
b
n
}
be two sequences of positive terms. Prove that
lim
a
n
b
n
6
lim
a
n
lim
b
n
,
when the product on the right is not of the form 0
· ∞
.
Proof
:
We distinguish 3 cases:
•
lim
a
n
=
∞
: Then by hypothesis
lim
b
n
>
0, so the righthand side is equal
to
∞
, and we have nothing to prove. By symmetry, the same happens
when
lim
b
n
=
∞
.
•
lim
a
n
= 0
: Then by hypothesis
lim
b
n
<
∞
and the righthand side is equal
to zero. So, we must prove that
lim
a
n
b
n
= 0, or equivalently lim
a
n
b
n
= 0
(indeed, if
lim
a
n
b
n
= 0 were true, then
0
6
lim
a
n
b
n
6
lim
a
n
b
n
= 0
,
so everything must equal zero). But this is true, since
a
n
b
n
6
Ma
n
,
for all
n
∈
N
, where
M
is the upper bound of the sequence
{
b
n
}
, which
exists since
lim
b
n
<
∞
. So 0
6
a
n
b
n
6
Ma
n
, for all
n
∈
N
, and by the
squeeze theorem we get that lim
a
n
b
n
= 0, as desired (again, we remind
that lim
a
n
= 0 holds since both
lim
a
n
and lim
a
n
are equal to zero). By
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 Winter '08
 hitrik
 Calculus, Topology

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