# What is the general term formula of this recurrence relation, if it exists?

$$a_n=-\frac{(a_{n-1} + 1)^2}{a_{n-1}+2}\quad \quad a_1 = -\frac 12$$

I transformed to:

$$a_n = -a_{n-1} - \frac{1}{a_{n-1} + 2}$$

$$a_{n-1} = -a_{n-2} - \frac{1}{a_{n-2} + 2}$$

...

And then

$$a_n - a_{n-1} + a_{n-2} -\cdots + \cdots - a_1$$

$$=$$

$$- a_{n-1} + a_{n-2} - ... + ... - \frac{1}{a_{n-1} + 2} + \frac{1}{a_{n-2} + 2} - ... + ...$$

Many terms cancel each other and:

$$a_n = - \frac{1}{a_{n-1} + 2} + \frac{1}{a_{n-2} + 2} - ... + ... + \frac{1}{a_{1} + 2} + a_1$$ (Here need to divide cases odd vs even)

And then I'm forever stuck. I can't find analytical form of the sum above

I try to get the value of the first few dozen terms to observe. I find nothing other than the terms are negative fractions that satisfy the recurrence relation.

Question is, is there a general term formula? How to find it?

• I heavily re-formatted your post. Please check to see if I introduced errors.
– lulu
Aug 4 '17 at 16:19
• It seems that $\lim_{n\to\infty} a_n = -1+\frac1{\sqrt2}$, although that might be only mildly useful to you. Aug 4 '17 at 17:22
• Probably useless, but if you set $a_n=\frac{b_n}{c_n}$, you can get two recurrence relations... Aug 4 '17 at 17:32
• @GregMartin: since $\lim a_n$ is clear by Banach fixed point Theorem, one also has $$a_n \approx -1+\frac{1}{\sqrt{2}}+C(5-4\sqrt{2})^n$$ for large values of $n$. That might be more useful. In order to have a closed form for a logistic-like recurrence we need to be pretty lucky, I do not know if that is the case here. To have some details on the original problem could be useful. Aug 5 '17 at 0:47
• @JackD'Aurizio The entirety of the original question is just "This is the recurrence relation, this is value of a(1). Find general term formula" Aug 5 '17 at 2:57

The recurrence is given by:

$$a_n=-\frac{(a_{n-1}+1)^2}{a_{n-1}+2}$$

Here we can separate both numerator and denominator as independent series. Let's create two new series, for them $c_n$ and $d_n$ respectively: $$a_{n-1}=\frac{c_{n-1}}{d_{n-1}}$$

So: $$a_n=-\frac{(\frac{c_{n-1}}{d_{n-1}}+1)^2}{\frac{c_{n-1}}{d_{n-1}}+2}\\ =\frac{-(c_{n-1}+d_{n-1})^2}{c_{n-1}+2d_{n-1}} =\frac{-c_{n-1}^2-2c_{n-1}d_{n-1}-d_{n-1}^2}{c_{n-1}+2d_{n-1}}\frac{1}{d_{n-1}}\\$$

Hence we can define the new sequences as: $$c_n=-c_{n-1}^2-2c_{n-1}d_{n-1}-d_{n-1}^2\\ d_n=c_{n-1}d_{n-1}+2d_{n-1}^2$$

Both are quadratic forms, very similar: $$c_n= \begin{bmatrix} c_{n-1} & d_{n-1} \end{bmatrix} \begin{bmatrix} p_{111} & p_{112}\\ p_{121} & p_{122} \end{bmatrix} \begin{bmatrix} c_{n-1}\\ d_{n-1} \end{bmatrix} \\ d_n= \begin{bmatrix} c_{n-1} & d_{n-1} \end{bmatrix} \begin{bmatrix} p_{211} & p_{212}\\ p_{221} & p_{222} \end{bmatrix} \begin{bmatrix} c_{n-1}\\ d_{n-1} \end{bmatrix}$$

with: $$\mathbf{p}_{1\cdot\cdot}=\begin{bmatrix} -1 & -1\\ -1 & -1 \end{bmatrix} \\ \mathbf{p}_{2\cdot\cdot}=\begin{bmatrix} 0 & 1/2\\ 1/2 & 2 \end{bmatrix}$$

Which can be summarized by renaming the numerator and denominator series as $\mathbf{x}_{1|n}$ and $\mathbf{x}_{2|n}$, and by using tensorial notation with implicit summation over repeated indexes, excepting the recurrence index: $$\mathbf{x}_{k_n}=\mathbf{x}_{i_{n-1}}\mathbf{p}_{k_ni_{n-1}j_{n-1}}\mathbf{x}_{j_{n-1}}$$

This recurrence, though tensorial, can be expressed up to the first term $\mathbf{x}_{\cdot|1}=[-1,2]$, where the tensor $\mathbf{p}$ appears $2^{n-1}-1$ times:

$$\mathbf{x}_{k_n}=\mathbf{x}_{i_{1}} \mathbf{p}_{i_{2}i_{1}j_{1}} \mathbf{x}_{j_{1}} \cdots \mathbf{p}_{k_ni_{n-1}j_{n-1}} \cdots \mathbf{x}_{i_{1}} \mathbf{p}_{j_{2}i_{1}j_{1}} \mathbf{x}_{j_{1}}$$

This could be depicted as a repetitive product of matrices, with $P$ appearing $2^{n-1}-1$ times: $$x_n=x_1 P x_1\cdots P \cdots x_1 P x_1$$

Though this expression is not closed and surely very far from what the OP wanted, this is not a recurrence, because it depends only on the first value of the sequence, and is a repetitive aplication of multidimensional products over the initial value.

• Hey, my comment came to good use! Aug 16 '17 at 19:31
• But i just realize now that you had it posted... So actually i never used it... So sad XD Aug 16 '17 at 19:32
• XD Whatevers! :P Aug 16 '17 at 19:33