Question

`lim_(n->oo)(((2n-1)!!)/(2n!!))`?

we have to prove that this converges to 0.

Answer 1

0

Explanation:

It is easy to see that

`(2n-1)!! = {(2n)!}/{2^n n!}`

and

`(2n)!! = 2^n n!`

So that the ratio is

`{(2n-1)!!}/{(2n)!!} = {(2n)!}/{2^{2n}(n!)^2}`

To evaluate the left hand side (let us call it `F_n`) you can use Stirling's approximation

`ln n! ~~ n ln n -n + 1/2 ln (2 pi n)`

so that

`ln F_n = ln {(2n)!} - 2n ln 2 -2 ln (n!)`
`~~ (2n) ln (2n) -2n + 1/2 ln (2 pi 2n)-2n ln 2 -2(n ln n -n+ 1/2 ln (2 pi n)) = ln ({sqrt{4pi n}}/{2 pi n})=ln (1/sqrt{pi n})`

Thus
`F_n ~~ 1/ sqrt{pi n}`

and so the limit as `n to infty` is zero.

You may feel uncomfortable with the use of the approximation here, especially since using the more common (and slightly less accurate) form of Stirling's approximation `ln n! ~~ n ln n - n` leads to the wrong answer. A more rigorous solution can be obtained by using a related inequality

`sqrt{2 pi n}(n/e)^n < n!< sqrt {e n}(n/e)^n`

(Source : https://en.m.wikipedia.org/wiki/Stirling%27s_approximation )

so that
# 0 < F_n < {sqrt {2e n}({2n}/e)^{2n}}/{2^{2n} (sqrt{2 pi n}(n/e)^n)^2}=sqrt (e/{2 pi^2 n})#

Answer 2

See below.

Explanation:

Assuming that

`((2k-1)!!)/(2k!!) = ((2k-1)(2k-3) cdots (3)(1))/((2k)(2k-2)cdots(4)(2))`

and calling `p_k = (2k-1)/(2k)` we have

`lim_(n->oo)((2n-1)!!)/(2n!!) = lim_(n->oo)prod_(k=1)^n p_k`

now making `p_k = 1+a_k = 1 - 1/(2k)` we have according with the considerations made in

https://socratic.org/questions/prove-that-this-converges-to-0-prod-k-1-n-lambdak-a-lambdak-b-0-a-b-lambda-0-n-1#549947

that `lim_(n->oo)prod_(k=1)^n p_k` diverges because

`lim_(n->oo)sum_(k=1)^n a_k = -1/2 sum_(k=1)^oo 1/k` diverges

Then as `0< p_k < 1` we have

`lim_(n->oo)((2n-1)!!)/(2n!!) =0`