502 Real Analysis
377.ForN≥2, define
aN=(
1 −
4
1
)(
1 −
4
9
)(
1 −
4
25
)
···
(
1 −
4
( 2 N− 1 )^2
)
.
The problem asks us to find limN→∞aN. The defining product foraNtelescopes as
follows:
aN=[(
1 −
2
1
)(
1 +
2
1
)][(
1 −
2
3
)(
1 +
2
3
)]
···
[(
1 −
2
2 N− 1
)(
1 +
2
2 N− 1
)]
=(− 1 · 3 )
(
1
3
·
5
3
)(
3
5
·
7
5
)
···
(
2 N− 3
2 N− 1
·
2 N+ 1
2 N− 1
)
=−
2 N+ 1
2 N− 1
.
Hence the infinite product is equal to
lim
N→∞
aN=−lim
N→∞2 N+ 1
2 N− 1
=− 1.
378.Define the sequence(aN)Nby
aN=∏N
n= 1(
1 +x^2
n)
.Note that( 1 −x)aNtelescopes as
( 1 −x)( 1 +x)( 1 +x^2 )( 1 +x^4 )···( 1 +x^2N
)
=( 1 −x^2 )( 1 +x^2 )( 1 +x^4 )···( 1 +x^2
N
)
=( 1 −x^4 )( 1 +x^4 )···( 1 +x^2
N
)
= ··· =( 1 −x^2N+ 1
).Hence( 1 −x)aN→1asN→∞, and therefore
∏n≥ 0(
1 +x^2n)
=1
1 −x.
379.LetPN =
∏N
n= 1 (^1 −xn
xn+ 1 ),N≥1. We want to examine the behavior ofPNas
N→∞. Using the recurrence relation we find that this product telescopes as
PN=
∏N
n= 1(
xn+ 1 −xn
xn+ 1)
=
∏N
n= 1nxn
xn+ 1=
N!
xN+ 1