# Non-linear recurrence relation $\frac{a_n - a_{n+1}}{1 + a_na_{n+1}} = \frac{1}{2n^2}$

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.

• Notice $\tan^{-1} a_n - \tan^{-1} a_{n+1}$ is known. If $a_n$ is a solution, then for any constant $\theta$, $\tan(\tan^{-1} a_n + \theta) = \frac{a_n + \tan\theta}{1 - a_n \tan\theta}$ is also a solution.... – achille hui Nov 20 '20 at 14:26
• @achillehui I get the gist, so we have infinitely many solutions! I was opting for that as well! – D. Petrov Nov 20 '20 at 14:29
• Have fun with wolframalpha.com/input/… – Claude Leibovici Nov 20 '20 at 15:20
• @ClaudeLeibovici Thank you for that idea! You had a little typo - missing 2 in the denominator there, but I got it from there. ;) Wolfram actually suggests an interesting solution in the real numbers: wolframalpha.com/input/… – D. Petrov Nov 20 '20 at 15:55
• I did not miss any $2$. WA solution is the same as mine : just make $c_1=\frac{1-a}{a-2}$. Interesting is to compare the alternate form to my approach. – Claude Leibovici Nov 21 '20 at 5:50

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 $$A_n=\frac{a-(a+1) n}{(a-1) n+1}$$

• Why do you assume the solution of this recurrence is a rational function of $n$? – Summand Nov 20 '20 at 14:05
• That's surely a more general form than mine, but it is still a restriction of a kind, as mentioned by @Summand Anyways, I'm really thankful for the effort! Just a little question, shouldn't we check that the denominator is not 0 before freeing ourselves from it? That is, add another condition that $A_nA_{n+1} \ne -1$? – D. Petrov Nov 20 '20 at 14:08
• @Summand. I know how to solve it using complex numbers. I started with the two solutions already available in the question and, if yo pay ttention, I wrote assume. – Claude Leibovici Nov 20 '20 at 14:15
• @D.Petrov. Have a look at my answer to @ summand. Cheers :-) – Claude Leibovici Nov 20 '20 at 14:17
• @ClaudeLeibovici would you mind explaining your strategy for solving the recurrence relation using complex numbers but without the hints (for example if we didn't know any of the solutions)? Thanks in advance! – Summand Nov 20 '20 at 14:18