Recurrence representation(s): $a(n+1)=a(n)(n-1/2)+o(1/n)$ and $a(n+1)=a(n)(n-1/2+o(1/n))$

Question

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.

Answer 1

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 $\displaystyle 1$.