## Task 1 - 231241

Let $\left(x_{n}\right)$ be the sequence of real numbers for which $x_{1}=1$ and it holds that:

$$
x_{n+1}=\frac{4 x_{n}^{2}+1}{5 x_{n}+1} \quad(n=1,2,3, \ldots)
$$

Investigate whether this sequence is convergent, and determine its limit if it exists.
Let $g$ be the positive solution of the equation $x^{2}+x-1=0$, so $g=\frac{\sqrt{5}-1}{2}$. (This is the reciprocal of the golden ratio $g=1 / \phi=\phi-1)$.

We will show that the sequence $\left(x_{n}\right)$ is monotonically decreasing and bounded below by $g$, and thus convergent. We will then show that $g$ is the limit of this sequence.

Lemma 1. For $x>g$, we have $x^{2}+x-1>0$.

Proof: $g$ is positive, and the other root of the equation is negative $\left(\frac{-\sqrt{5}-1}{2}\right)$, so $g$ is the larger of the two roots. To the right of the larger root, a quadratic function with a positive coefficient for $x^{2}$ only takes positive values.

Lemma 2. For $x>g>0$, we have $\frac{4 x^{2}+1}{5 x+1}>g$.

Proof: Let $d:=x-g>0$. Then $3 g d+4 d^{2}>0$ equivalent transformations yield

$$
4 g^{2}+8 g d+4 d^{2}+1>4 g^{2}+5 g d+1=4 g^{2}+5 g d+\left(g^{2}+g\right)
$$

(by definition of $g$) and further

$$
4(g+d)^{2}+1>5 g^{2}+5 g d+g=g(5 g+5 d+1)
$$

or $4 x^{2}+1>g(5 x+1)$. Division by $5 x+1>0$ gives the claim.

Lemma 3. For $x>g>0$, we have $\frac{4 x^{2}+1}{5 x+1}<x$.

Proof: Since $x>g$, we have $x^{2}+x-1>0$, which implies $x^{2}>1-x$. Multiplication by 4 gives $4 x^{2}>4-4 x$. Adding $x$ to both sides yields $4 x^{2}+x>4-3 x$. Adding $x^{2}$ to both sides gives $5 x^{2}+x=x(5 x+1)>4 x^{2}+1$. Division by $5 x+1>0$ gives the claim.

With Lemma 2 and 3, we can now prove by induction that $g\frac{\sqrt{5}-1}{2}=g$.

Base case: $x_1 = 1 > g$.

Inductive step: Assume $x_{n}>g$. By Lemma 2 and 3, we have $g0$ and bounded below by $g$. Therefore, $\left(x_{n}\right)$ is convergent. Let $x>0$ be the limit of $\left(x_{n}\right)$, then $x$ must satisfy the equation $x=\frac{4 x^{2}+1}{5 x+1}$ $(*)$. Multiplying by $5 x+1$ and simplifying, we get: $x^{2}+x-1=0$. Thus, $x$ is the positive root of $x^{2}+x-1$, which is precisely $g$.

(*) If $x_{n+1}=\frac{4 x_{n}^{2}+1}{5 x_{n}+1}$ for all $n$ and $\left(x_{n}\right)$ converges to $x$, then

$$
x=\lim _{n \rightarrow \infty} x_{n+1}=\lim _{n \rightarrow \infty} \frac{4 x_{n}^{2}+1}{5 x_{n}+1}=\frac{\lim _{n \rightarrow \infty} 4 x_{n}^{2}+1}{\lim _{n \rightarrow \infty} 5 x_{n}+1}=\frac{4\left(\lim _{n \rightarrow \infty} x_{n}\right)^{2}+1}{5 \lim _{n \rightarrow \infty} x_{n}+1}=\frac{4 x^{2}+1}{5 x+1}
$$