Recurrence representation(s): $a(n+1)=a(n)(n-1/2)+o(1/n)$ and $a(n+1)=a(n)(n-1/2+o(1/n))$ I know that the recurrence $\displaystyle a(n+1)=a(n)(n-1/2)$ can be represented like $\displaystyle \frac{(2n-1)!!}{2^n}$
Actually the initial recurrence is slightly different:
$$\displaystyle a(n+1)=a(n)(n-1/2)+o(1/n)$$ 
I am looking  for the new representation of it in terms of $\displaystyle \frac{(2n-1)!!}{2^n}$.
Probably the question has to be refined. The recurrence is very close to the property of Kendall-Mann numbers http://oeis.org/A181609
that is why I am trying to find an interval of $\displaystyle n$'s when the $\displaystyle \frac{(2n-1)!!}{2^n}$ or similar representation works fine.
I understand that $\displaystyle o(1/n)$ is quite big error for the representation.
I am also interested in the case the recurrence is
$$\displaystyle a(n+1)=a(n)(n-1/2+o(1/n))$$ 
which is closer to the recurrence satisfied by the Kendall-Mann numbers.
 A: 
For the recurrence $\displaystyle a_{n+1} = (n - 1/2) a_n + o(1/n)$

If you set $\displaystyle a_n = \frac{(2n-1)!!}{2^n} - e_n$,
you get $\displaystyle e_{n+1} - e_n = o\left(\frac{1}{n}\right)$ and thus $e_n = o(\log n)$.
Using Stirling's approximation, I believe we have that
$\displaystyle \frac{(2n-1)!!}{2^n} \approx \frac{Kn!}{\sqrt{n}}$
Thus we get
$$\lim_{n \to \infty} \frac{2^n a_n}{(2n-1)!!} = 1 $$

Now if the recurrence was 
  $\displaystyle a_{n+1} = a_{n} (n - 1/2 + o(1/n))$

Set $\displaystyle a_{n+1} = e_n\frac{(2n-1)!!}{2^n}$
We get
$\displaystyle \frac{e_{n+1}}{e_n} = 1 + o(1/n^2)$
And thus
$\displaystyle  e_n = e_1 \prod_{k=1}^{n-1} (1 + o(1/k^2))$
Now using $1 + x \le e^x$ we get that
$\displaystyle e_n = \mathcal{O}(1)$
and we can easily show that $\displaystyle e_n$ is convergent.
Thus 
$$a_n \sim \frac{C(2n-1)!!}{2^n}$$
for some constant $\displaystyle C = \lim_{n \to \infty} e_n$ .
We could chose your $\displaystyle f(n) = o(1/n)$ as we like, to make this constant different from $\displaystyle 1$.
