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The evaluation of b
(
n
)
is strictly related to the following Wallis' product for
π /
2(a
consequence of Euler's jewel given in Sect. 5.4.3):
4 n 2
4 n 2
4 i 2
4 i 2
= n > 0
i
π /
2
1 =
lim
n
(7.5)
1
=
1
,
n
2 =
2
1 ·
2
3 ·
4
3 ·
4
5 ·
6
5 ·
6
7 ... ·
2 n
2 n
2 n
2 n
1 ·
1 ...
(7.6)
+
if we denote by
(
2 n
)
!! the product of the even numbers equal to or less than 2 n ,we
get:
2 n
(
2 n
)
!!
=
·
n !
(7.7)
moreover, if we denote by
(
2 n
+
1
)
!! and by
(
2 n
1
)
!! the product of the odd num-
bers equal to or less than 2 n
+
1and2 n
1 respectively, we get:
2 n
(
2 n
1
)
!!
=(
2 n
)
!
/ (
·
n !
)
(7.8)
2 n
(
2 n
+
1
)
!!
=(
2 n
+
1
) · (
2 n
)
!
/ (
·
n !
)
(7.9)
therefore, Eq. (7.6) becomes:
(
2 n
)
!!
· (
2 n
)
!!
π /
2
n
!!
whence, according to Eqs. (7.7), (7.8), and (7.9), it follows that:
(
2 n
+
1
)
!!
· (
2 n
1
)
2 n n !
2 n n !
·
π /
2
n
(
2 n
+
1
) · (
2 n
)
!
/ (
2 n
·
n !
) · (
2 n
)
!
/ (
2 n
·
n !
)
that is:
2 4 n
4
2 n =
(
n !
)
) .
2
((
2 n
)
!
)
(
2 n
+
1
2 , b
2
Now, let us consider the terms
(
b
(
n
))
(
2 n
)
and evaluate the ratio
(
b
(
n
))
/
b
(
2 n
)
.
We h ave :
2 e 2 n
n 2 n
= (
)
n !
2
(
b
(
n
))
! e 2 n
)= (
2 n
)
(
b
2 n
(
2 n
)
2 n
2
2 e 2 n
n 2 n
2 n
2
2 2 n n 2 n
(
2 2 n
2
(
b
(
n
))
) = (
n !
)
(
2 n
)
! e 2 n = (
n !
)
(
n !
)
! =
n 2 n
b
(
2 n
(
2 n
)
2 n
)
(
2 n
)
!
1
/
2 by
1
/
2 , becomes:
(
+
)
(
)
which, by approximating
2 n
1
2 n
2
n π
(
b
(
n
))
1
/
2
1
/
2
) =(
2 n
+
1
)
( π /
2
)
n
.
b
(
2 n
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