converge........

[mina_world]

hello.....doctor~

i want to show that

lim {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)} = 0
n->00

but i can't....

maybe.....sandwich ??

i know that 3*5*7*...*(2n-1) > 2*4*6*...*(2n-2).

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

> {3*4*6*....*(2n-2)} / {2*4*6*....*(2n)} = 1/(2n).

but {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)} < ???

um......help me, please~

thank you very much.

[Julien Santini]

> i want to show that
>
> lim {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)} = 0
> n->00
>
> but i can't....

= (2n)!/(n!2^n)^2 ~ (2n)^(2n)exp(-2n)sqrt(4Pi*n)/((2n)^(2n)exp(-2n)2Pi*n) ~
1/sqrt(Pi*n) (or something like that) which converges to 0.

[Don Coppersmith]

On 12 Aug 2004, mina_world wrote:
>hello.....doctor~

>
>
>i want to show that
>
>lim {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)} = 0
>n->00

Consider the square of that quantity,
(1*1*3*3*5*5...*(2n-1)*(2n-1)} / {2*2*4*4*...*(2n)*(2n)}
Take pieces like
{(2k-1)*(2k+1)} / {(2k)*2k)} < 1,
e.g. (3*5)/(4*4)=15/16<1.
Take out as many of those as you can; what's left?

[José Carlos Santos]

mina_world wrote:

> i want to show that

>

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

> n->00

>

> but i can't....

Prove the the sequence of the logarithms has limit -oo.

Best regards,

Jose Carlos Santos

[The World Wide Wade]

In article <cfg30n$het$1...@news.hananet.net>,
"mina_world" <mina_...@hanmail.net> wrote:

> i want to show that
>
> lim {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)} = 0
> n->00

Show the log of this -> -oo. The inequality ln(1 + x) <= x will be useful.

[David McAnally]

dco...@us.ibm.com (Don Coppersmith) writes:

This gives what is wanted, certainly. But there is another comment that
I would like to make about these factors which is completely separate to
the point that you are making here.

The infinite product whose factors are those as you listed here has a
positive (and therefore nonzero) limit. This is not surprising since the
product is conditionally convergent, and so the order of the factors can
be rearranged to give the product any nonnegative value that you would
like (even infinity).

David

And all dared to brave unknown terrors, to do mighty deeds,
to boldly split infinitives that no man had split before -
and thus was the Empire forged.

-----

[Vitaliy Kushnirevych]

mina_world wrote:
> hello.....doctor~
>
>
> i want to show that
>
> lim {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)} = 0
> n->00
>
> but i can't....
>
>

The easiest (imho) way to show is to use almost evident inequality:
1/(n+1) < ln (1 + 1/n) < 1/n

Then
ln A_n=ln( {2*4*...*(2n)} /{1*3*5*...*(2n-1)} =
ln(1+1)+ln(1+1/3)+ln(1+1/5)+...+ ln(1+1/(2n-1)

An we have
1/2 + 1/4 +...+ 1/(2n) < ln A_n
Since the left hand side as n -> oo diverge to +oo, we have the result.

[Zdislav V. Kovarik]

On Fri, 13 Aug 2004, mina_world wrote:

> hello.....doctor~
>
>
> i want to show that
>
> lim {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)} = 0
> n->00
>
> but i can't....

It requires a little trick, if we want to keep it elementary:

Let a(n) = {1*3*5*....*(2n-1)} / {2*4*6*....*(2n)}

(as in your problem);

and consider c(n) = a(n) * (2*n + 1)^(1/2).

With some patience, you can prove that {c(n)} is a decreasing sequence
of positive numbers. How? Show that

(c(n+1) / c(n))^2 < 1 .

So, c(n) <= c(1), hence

0 < a(n) <= c(1) / (2*n+1)^(1/2)

and you have your squeeze.

Cheers, ZVK(Slavek).