
2. We consider number sequences $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{n+1}=\frac{a_{n}+a_{1}}{a_{n}+1}$ holds for all $n \geqslant 1$.

(a) Suppose that $a_{1}=-3$. Compute $a_{2020}$.

(b) Suppose that $a_{1}=2$. Prove that $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$ holds for all $n \geqslant 2$.

## 2. Version for klas 5 \& klas 4 and below

(a) If $a_{1}=-3$, we have $a_{2}=\frac{a_{1}+a_{1}}{a_{1}+1}=\frac{-6}{-2}=3$. Next, we find $a_{3}=\frac{a_{2}+a_{1}}{a_{2}+1}=\frac{0}{4}=0$. Then, we have $a_{4}=\frac{a_{3}+a_{1}}{a_{3}+1}=\frac{-3}{1}=-3$. We see that $a_{4}=a_{1}$. Since $a_{n+1}$ only depends on $a_{n}$ and $a_{1}$, we see that $a_{5}=a_{2}=3, a_{6}=a_{3}=0, a_{7}=a_{4}=-3$, et cetera. In other words: the sequence is periodic with period 3 , and we see that

$$
a_{2020}=a_{2017}=a_{2014}=\cdots=a_{4}=a_{1}=-3 .
$$

(b) We will prove the statement by induction on $n$. We start with the induction basis $n=2$. We calculate $a_{2}=\frac{a_{1}+a_{1}}{a_{1}+1}=\frac{4}{3}$ and see that indeed $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$ holds.

Now suppose that for certain $n=k \geqslant 2$ we have proven that $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$. We will show that these inequalities hold for $n=k+1$ as well.

We first observe that

$$
a_{k+1}=\frac{a_{k}+2}{a_{k}+1}=1+\frac{1}{a_{k}+1} .
$$

From the induction hypothesis $\left(\frac{4}{3} \leqslant a_{k} \leqslant \frac{3}{2}\right)$ it follows that $\frac{7}{3} \leqslant a_{k}+1 \leqslant \frac{5}{2}$, and hence that

$$
\frac{2}{5} \leqslant \frac{1}{a_{k}+1} \leqslant \frac{3}{7} .
$$

By adding 1 to all parts of these inequalities, we find

$$
\frac{7}{5} \leqslant 1+\frac{1}{a_{k}+1}=a_{k+1} \leqslant \frac{10}{7} .
$$

Since $\frac{4}{3} \leqslant \frac{7}{5}$ and $\frac{10}{7} \leqslant \frac{3}{2}$, we see that $\frac{4}{3} \leqslant a_{k+1} \leqslant \frac{3}{2}$, completing the induction step.

