In the last steps of finding the complete solution of a linear differential equation by a power series, I got stuck on finding the closed formula for the following recurrent relation:
$$B_n = B_{n-1} \times {2n-5\over 2n^2 - n}$$
Can you please explain how can I get the $B_n$ expression that depends only on $n$ and $B_0$?
What I got so far was:
$$B_1\,B_2\,B_3\dots B_n = -3B_0 \times{-1\over6}B_1 \times {1\over15}B_2 \times \dots\times B_{n-1} {2n-5\over 2n^2 - n}$$
Dividing everything by $B_1\, B_2\, B_3 \dots B_{n-1}$, I got:
$$B_n = B_0 \times {2(1)-5\over 2(1)^2 - 1} \times {2(2)-5\over 2(2)^2 - 2} \times {2(3)-5\over 2(3)^2 - 3} \times {2(4)-5\over 2(4)^2 - 4} \times \dots \times {2n-5\over 2n^2 - n}$$
But I don't know what to do from now on...
Writing the denominator as $n(2n-1)$ as correctly suggested in the comments, you can focus on the simplified recurrence relation given by $B_n=B_{n-1}(2n-5)/(2n-1)$. Finding a closed formula for this, we will obtain a closed formula even for the recurrence relation described in the question, simply adding a factorial in the denominator.
It is not difficult to see that the simplified recurrence relation, for $n\geq{1}$, can be written as $$B_n=B_0\frac{-3*-1*1*3*5....*(2n-5)}{1*3*5*7*9....*(2n-1)}=B_0\frac{3}{(2n-3)(2n-1)}$$
The closed formula is therefore $$B_n=B_0\frac{3}{n!(2n-3)(2n-1)}$$.