A sequence is represented by a0, a1, a2, a3, ..., an, where an=f(n) is a possible function for determining an generally. But this is only one of a number of ways of defining an. For example, a Fibonacci series defines an=an-2+an-1 and an cannot be defined (exactly) by using a function of n, although there is a formula for an approximation of an.
But if an is given as a function of n, then an+1 can easily be found by substitution.
We can also find an+1-an and thereby establish a relationship between consecutive terms. A simple example is, given an=(2n+1)/(2n-1), the series would be:
-1, 3, 5/3, 7/5, 9/7, ...
an+1=(2(n+1)+1)/(2(n+1)-1)=(2n+3)/(2n+1);
an+1-an=(2n+3)/(2n+1)-(2n+1)/(2n-1)=[(2n+3)(2n-1)-(2n+1)2]/(4n2-1),
an+1-an=(4n2+4n-3-4n2-4n-1)/(4n2-1)=-4/(4n2-1), so:
an+1=an-4/(4n2-1).
Now take one of your examples. an=(1)(3)(5)...(2n+1)/[(2)(4)(6)...(2n)].
an+1=(1)(3)(5)...(2n+3)/[(2)(4)(6)...(2n+2)].
an+1-an=(1)(3)(5)...(2n+3)/[(2)(4)(6)...(2n+2)]-(1)(3)(5)...(2n+1)/[(2)(4)(6)...(2n)],
an+1=an+[(2n+3)/(2n+2)]{(1)(3)(5)...(2n+1)/[(2)(4)(6)...(2n)]}. Although it's complicated, there is a clear function of n which links successive terms of the series. This series could also be expressed:
Given a0=1, an+1=(2n+3)/(2n+2)an, generating the series an=1, 3/2, 15/8, 105/48=35/16, ... Note that this is different from an+1=an+f(n), which is a sum, not a product.
I think the problem you have observed is, not how to calculate xn+1, but how to calculate an as f(n), a function of n. If and when you can do so, then you will be able to calculate another function of n, g(n), which relates successive terms in the series.