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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...

lvella
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    Factor the denominator: $2n^2-n = n(2n-1)$. The factors $n$ produce an $n!$ in the denominator of $B_n$, What remains is $\frac{2n-5}{2n-1}$. Look at it, take some instrument that helps looking at far away things, if necessary. – Daniel Fischer Jul 02 '14 at 21:54
  • Some instrument? Something like limit? If so, I don't even know how to write it down... – lvella Jul 02 '14 at 22:54
  • If you work out the first 10 or so values of $1/(n!B_n)$, you should see the pattern that will give you the general formula (especially if you plot them). – Greg Martin Jul 02 '14 at 23:00
  • This was the instrument Daniel Fischer was talking about: http://images.hayneedle.com/mgen/master:ZHUL053.jpg – JimmyK4542 Jul 03 '14 at 02:46

1 Answers1

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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 $$B_n=B_0\frac{3}{n!(2n-3)(2n-1)}$$.

Anatoly
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