Is there a way to solve this non-linear recurrence relation? While exploring some ideas of my own, I came across the following recurrence relation:
$$a_{n+1}=\frac{1}{4(\frac{1}{2n}-a_n)(2n+1)^2}$$
for $n\ge1$ with $a_1$ arbitrary.
I have no experience with non-linear recurrence relations and I realise that they are generally considered to be intractable. Is this one also?
I managed to guess the solution to a much simpler one that also came up. For this sequence, it turns out that $a_1 = \frac{1}{4a_0}$. In case anyone is better at guessing than me, here are the first few terms of this sequence with $a_0 = 1$. I think it might also help to consider $b_{n+1} = (2n+1)^2a_{n+1}.$
$$\begin{align*} a_2 = \frac{1}{9}, \quad b_2 = 1 \end{align*}$$
$$a_3 = \frac{9}{125}, \quad b_3 = \frac{9}{5}$$
$$a_4 = \frac{375}{6958}, \quad b_2 = \frac{375}{142}$$
At which point, my calculator no longer can write the answer as a fraction.
Also, by construction, I can promise that this sequence tends to zero, if that helps, and it is probably strictly decreasing.
 A: There is a classical method called fixed point method.
Firstly, let
$$
x_n:=\frac{1+\sqrt{\frac{4n+1}{(2n+1)^2}}}{4n}\quad \text{and}\quad y_n:=\frac{1-\sqrt{\frac{4n+1}{(2n+1)^2}}}{4n}.
$$
It is easy to see that $x_n,y_n$ are the two solutions of the following equation:
\begin{align*}
x=\frac{1}{4(\frac{1}{2n}-x)(2n+1)^2}.
\end{align*}
Subtract $x=x_n,y_n$ from both sides of the recurrence relation to get
\begin{align*}
a_{n+1}-x=\frac{1}{4(\frac{1}{2n}-a_n)(2n+1)^2}-x
\end{align*}
\begin{align}
\Rightarrow 4(a_{n+1}-x)=\frac{(a_n-x)x}{(\frac{1}{2n}-a_n)},\quad x=x_n\text{ or } y_n.\tag{A}
\end{align}
Denote by
\begin{align*}
b_n:=\frac{a_n-x_n}{a_n-y_n}.
\end{align*}
By (A), we have
\begin{align*}
4b_{n+1}=\frac{x_n}{y_n}b_n\Rightarrow b_n=\frac{1}{4^{n-1}}\Pi_{i=2}^n \frac{x_i}{y_i}b_1=\frac{1}{4^{n-1}}\Pi_{i=2}^n \frac{x_i}{y_i}\frac{a_1-x_1}{a_1-y_1}\quad\text{for $n\ge2$}.
\end{align*}
Notice that
\begin{align*}
a_n=\frac{b_ny_n-x_n}{b_n-1},
\end{align*}
which gives the formula of $a_n$.
This method looks a bit complex, and I haven't check if there is any error, like $\frac{1}{0}$, in my formulation. Hope this helps you!
