Problem: I need to find all sequences $\{a_n\}_{n=1}^\infty$ that satisfy the following non-linear recurrence relation: $$\frac{a_n - a_{n+1}}{1 + a_na_{n+1}} = \frac{1}{2n^2}$$
What would the general approach be, if there is any? I couldn't find any certain literature on that topic, so pointing me in some direction would be really helpful!
Motivation: As you probably recognize, this is namely the difference of arctangents formula: $$arctan(a) - arctan(b) = arctan\left(\frac{a - b}{1 + ab}\right)$$
I was initially confronted with the following series: $$\sum_{n=1}^\infty arctan\left(\frac{1}{2n^2}\right)$$ and my initial intuition was that it's a telescopic series. So the approach was namely the named above - to find a sequence $a_n$ such that I can present $arctan\left(\frac{1}{2n^2}\right) = arctan(a_n) - arctan(a_{n+1})$ and thus the general solution for the value of the series would be: $$\sum_{n=1}^\infty arctan\left(\frac{1}{2n^2}\right) = arctan(a_1) - \lim_{N\to\infty}arctan(a_{N+1})$$ Of course, I searched the web for this certain problem and people were suggesting various options for $a_n$, the most common of which were:
- $a_n = \frac{1}{2n-1}$
- $a_n = 1 - 2n$
Both satisfy the mentioned recurrence relation and lead to a satisfying result for the series, namely $\sum_{n=1}^\infty arctan\left(\frac{1}{2n^2}\right) = \frac{\pi}{4}$
Anyways, I was simply not satisfied with the explanation that this series requires some "clever rewriting" as a telescopic one. I wanted to build some stronger intuition as to how to find such rewritings, when I have the idea that a series might be telescopic. I just couldn't resist wondering whether these are for some reason the only two possible sequences, whether there are more possible ones or they are maybe infinitely many? Obviously, all boils down to generally solving the non-linear recurrence relation.
My progress:
I would like also share my approach to this problem, which only leads to the solution $a_n = 1 - 2n$ since I'm making a restriction at the beginning in order to make any reasonable progress. What I'm basically doing is expressing: $$\frac{a_n - a_{n+1}}{1 + a_na_{n+1}} = \frac{d}{2dn^2}$$
And artificially building the following system (which, naturally, loses solutions, so I'm not happy...): $$\begin{array}{|l}a_n - a_{n+1} = d \\ 1 + a_na_{n+1} = 2dn^2\end{array}$$
After some transformations and case distinction, one is able to obtain $a_n = 1 - 2n$. But the first solution I mentioned simply doesn't fit this form, so I need something more general as an approach to the recurrence relation.
Many thanks in advance!
This kind of recurrence relations are quite difficult to solve. If my memory is good (not too sure at my age !), I think I saw them solve using complexes.
Anyway, taking into account what you had as solution, let us try for $$\frac{A_n - A_{n+1}}{1 + A_n\,A_{n+1}} -\frac{1}{2n^2}=0$$ Reduce to common denominator
$$2 n^2 A_n-2 n^2 A_{n+1}-A_n A_{n+1}-1=0$$
Now, assume that $A_n=\frac{a+bn}{1+cn}$; replace and reduce to same denomiantors everywhere to arrive at $$-\left(a^2+a b+c+1\right)-n \left(2 a b+b^2+c^2+2 c\right)+n^2 \left(2 a c-b^2-2 b-c^2\right)=0$$ Now, we need to cancel all these coefficients.
The first one gives $c=-a^2-ab-1$. Replace to obtain $$\left(a^2+1\right)(a+b+1) \big[ (a+b-1) n+ (a+b+1) n^2\big]=0$$ So $b=-a-1$ and $c=a-1$
So $$A_n=\frac{a-(a+1) n}{(a-1) n+1}$$