Final Round

a. First, we determine for what starting values $a_1>0$ the inequalities $\frac{4}{3}\le a_2\le \frac{3}{2}$ hold. Then, we will prove that for those starting values, the inequalities $\frac{4}{3}\le a_n\le \frac{3}{2}$ are also valid for all $n\ge 2$.

First, we observe that $a_2=\frac{a_1+2}{a_1+1}$ and that the denominator, $a_1+1$, is positive (since $a_1>0$). The inequality $$\frac{4}{3}\le a_2=\frac{a_1+2}{a_1+1}\le \frac{3}{2},$$ is therefore equivalent to the inequality $$\frac{4}{3}(a_1+1)\le a_1+2\le \frac{3}{2}(a_1+1),$$ as we can multiply all parts in the inequality by the positive number $a_1+1$. Subtracting $a_1+2$ from all parts of the inequality, we see that this is equivalent to $$\frac{1}{3}{a}_1-\frac{2}{3}\le 0\le \frac{1}{2}{a}_1-\frac{1}{2}.$$ We therefore need to have $\frac{1}{3}{a}_1\le \frac{2}{3}$ (i.e. $a_1\le 2$), and $\frac{1}{2}\le \frac{1}{2}{a}_1$ (i.e. $1\le a_1\le 2$). The starting value $a_1$ must therefore satisfy $1\le a_1\le 2$.

Now suppose that $1\le a_1\le 2$, so that $a_2$ satisfies $\frac{4}{3}\le a_2\le \frac{3}{2}$. Looking at $a_3$, we see that $a_3=\frac{a_2+2}{a_2+1}$. That is the same expression as for $a_2$, only with $a_1$ replaced by $a_2$. Since $a_2$ also satisfies $1\le a_2\le 2$, the same argument now shows that $\frac{4}{3}\le a_3\le \frac{3}{2}$.

We can repeat the same argument to show this for $a_4$, $a_5$, etcetera. Hence, we find that $\frac{4}{3}\le a_n\le \frac{3}{2}$ holds for all $n\ge 2$. The formal proof is done using induction: the induction basis $n=2$ has been shown above. For the induction step, see the solution of part (b) of the version for klas 5 & klas 4 and below. The result is that all inequalities hold if and only if $1\le a_1\le 2$.

b. Let's start by computing the first few numbers of the sequence in terms of $a_1$. We see that $$a_2=\frac{a_1-3}{a_1+1}$$ and $$\begin{array}{rl}{a}_3& =\frac{a_2-3}{a_2+1}=\frac{\frac{a_1-3}{a_1+1}-3}{\frac{a_1-3}{a_1+1}+1}=\frac{a_1-3-3(a_1+1)}{a_1-3+(a_1+1)}\\ & =\frac{-2a_1-6}{2a_1-2}=\frac{-a_1-3}{a_1-1}.\end{array}$$ Here, it is important that we do not divide by zero, that is, $a_1\ne -1$ and $a_2\ne -1$. The first inequality follows directly from the assumption. For the second inequality we consider when $a_2=-1$ holds. This is the case if and only if $a_1-3=-(a_1+1)$, if and only if $a_1=1$. Since we assumed that $a_1\ne 1$, we see that $a_2\ne -1$. The next number in the sequence is $$a_4=\frac{a_3-3}{a_3+1}=\frac{\frac{-a_1-3}{a_1-1}-3}{\frac{-a_1-3}{a_1-1}+1}=\frac{-a_1-3-3(a_1-1)}{-a_1-3+(a_1-1)}=\frac{-4a_1}{-4}=a_1.$$ Again, we are not dividing by zero since $a_3=-1$ only holds when $-a_1-3=-a_1+1$, which is never the case.

We see that $a_4=a_1$. Since $a_{n+1}$ only depends on $a_n$, we see that $a_5=a_2$, $a_6=a_3$, $a_7=a_4$, etcetera. In other words: the sequence is periodic with period 3, and we see that $$a_{2020}=a_{2017}=a_{2014}=\cdots =a_4=a_1.$$ To conclude: indeed we have $a_{2020}=a_1$ for all starting values $a_1$ unequal to 1 and $-1$.