Re: Analysis with limit before.

In article <g1ov5a$sqs$1@xxxxxxxxxxxxxxxxxxxxx>,
"mina_world" <mina_world@xxxxxxxxxxx> wrote:
lim {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)} = 0
n->00

http://groups.google.com/group/sci.math/browse_frm/thread/4022a467733f198b

I already asked before.

(1)
1/2 < 2/3
3/4 < 4/5
5/6 < 6/7
...
(2n-1)/(2n) < (2n)/(2n+1)

so,
[(1/2).(3/4).(5/6)...(2n-1 / 2n)]^2
< (1/2).(3/4).(5/6)...(2n-1 / 2n)]*[(2/3).(4/5).(6/7)...(2n / 2n+1)]
= 1 / (2n+1)

so,
(1/2).(3/4).(5/6)...(2n-1 / 2n) < Sqrt[1 / (2n+1)]


(2)
1/2 = 1/2
2/3 < 3/4
4/5 < 5/6
...
(2n-2)/(2n-1) < (2n-1)/(2n)

so,
[(1/2).(3/4).(5/6)...(2n-1 / 2n)]^2
(1/2).(3/4).(5/6)...(2n-1 / 2n)]*[(1/2).(2/3).(4/5)...(2n-2 / 2n-1)]
= 1 / (4n)

so,
(1/2).(3/4).(5/6)...(2n-1 / 2n) < Sqrt[1 / (4n)]

Thus, {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)} --> 0

This is true.

However, using the Euler-Maclaurin Sum Formula, we can show even
more. The Euler-Maclaurin Sum Formula says that

---
> log(1-1/(2k))
---
k=1

= C + n log(1-1/(2n)) - 1/2 log(n)
+ 1/12 (1/(n-1/2) - 1/n)) - 1/360 (1/(n-1/2)^3 - 1/n^3))
+ 1/1260 ((1/(n-1/2)^5 - 1/n^5)) - 1/1680 ((1/(n-1/2)^7 - 1/n^7))
+ 1/1188 ((1/(n-1/2)^9 - 1/n^9)) - ...

Using n = 10, we get 12 places of precision, so we can compute that
C = -0.072364942924... This allows us to compute the following limit

n
--- 2k-1
lim sqrt(n) | | ----
n->oo k=1 2k

which is exp(C-1/2) = 0.564189583548...

For large n, your product is approximately 0.564189583548/sqrt(n).

Rob Johnson <rob@xxxxxxxxxxxxxx>
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