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

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?

  • 2
    $\begingroup$ I heavily re-formatted your post. Please check to see if I introduced errors. $\endgroup$
    – lulu
    Aug 4, 2017 at 16:19
  • $\begingroup$ It seems that $\lim_{n\to\infty} a_n = -1+\frac1{\sqrt2}$, although that might be only mildly useful to you. $\endgroup$ Aug 4, 2017 at 17:22
  • $\begingroup$ Probably useless, but if you set $a_n=\frac{b_n}{c_n}$, you can get two recurrence relations... $\endgroup$ Aug 4, 2017 at 17:32
  • 1
    $\begingroup$ @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. $\endgroup$ Aug 5, 2017 at 0:47
  • $\begingroup$ @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" $\endgroup$
    – Boyang
    Aug 5, 2017 at 2:57

1 Answer 1


This is an incomplete answer.

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.

  • $\begingroup$ Hey, my comment came to good use! $\endgroup$ Aug 16, 2017 at 19:31
  • $\begingroup$ But i just realize now that you had it posted... So actually i never used it... So sad XD $\endgroup$
    – Brethlosze
    Aug 16, 2017 at 19:32
  • $\begingroup$ XD Whatevers! :P $\endgroup$ Aug 16, 2017 at 19:33

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