Strange Sequence

From the theory of linear recurrence relations , since the characteristic equation is $m^2 +m - 2 = (m+2)(m-1)$ , hence the recurrence relation has the form $a_i = A (-2)^i + B(1)^i$ .

If $A \neq 0$ , then we eventually would have a large enough $i$ such that $A ( -2)^i < - |B|$ , which would make $a_i = A(-2)^i + B < 0$ . Hence, we have $A = 0$ and so $a_i = B$ is the constant sequence.

Thus, $a_{2017} = a_0 = 1$ .

This question doesn't seem to have enough information since we don't even know what $a_1$ is! However, we will soon find that we have all the information we need. Remember, the sequence is defined for positive reals. For now, let $a_1=x$ . We have

$\begin{aligned} a_0 = 1 &\\ a_1 = x &\implies x>0\\ a_2 = 2 - x &\implies x < 2\\ a_3 = 3x - 2 &\implies x > \frac{2}{3}\\ a_4 = 6 - 5x &\implies x < \frac {6}{5}\\ a_5 = 11x - 10 &\implies x > \frac{10}{11}\\ \end{aligned}$

It seems that as the terms progress further, $a_1$ is being squeezed between two fractions towards 1. Let's try and prove that $a_1=1$ .

Firstly, let us set $x=0$ so we get a new sequence $b_0=1, b_1=0, b_2 = 2, b_3=-2, b_4=6, b_5=-10, \ldots$ . Notice that from $b_2$ onwards, the sequence alternates in sign, with $|b_{n+2}| > |b_n|$ . Let's try and prove this conjecture via induction.

Let $b_{2n}=p, b_{2n+1}=-q$ , where $n, p, q$ are positive integers. If $n=1$ , we have $p=q=2$ , which alternate in sign, as per the induction hypothesis.

If $n=k$ , we have $b_{2k+2}=2r-s>r>0$ , and $b_{2k+3} = -(2r+3s)<0$ , with $|-(2r+3s)|>s$

From this, we get $b_{2k+2}>b_{2k}>0$ and $b_{2k+3}<b_{2k+1}<0$ , so our conjecture must be true by induction.

Notice also that the difference between the absolute value of the coefficients of $x$ and the absolute value of the constant term in each term of the recurrence is 1. We will now show this to be true. If we substitute $x=1$ , we get $a_n=1 \forall n \in \mathbb{N}_0$ . Thus, our observation must be true. Therefore,

$\begin{aligned} a_{2n} = b_{2n} - (b_{2n} - 1)x &\implies x < \dfrac{b_{2n}}{b_{2n}-1}\\ a_{2n+1} = b_{2n+1} - (b_{2n+1} - 1) x &\implies x > \dfrac {b_{2n+1}}{b_{2n+1}-1}\\ \end{aligned}$

Since both $|b_{2n}|$ and $|b_{2n+1}|$ are strictly increasing with $n$ , the limit as $n$ tends to $\infty$ of both $\frac{b_{2n}}{b_{2n}-1}$ and $\frac {b_{2n+1}}{b_{2n+1}-1}$ is 1.

Therefore, by Squeeze Theorem, we must have $x=1$ , so $a_n=1 \forall n \in \mathbb{N}_0$ . Thus, the sum of all values of $a_{2017}$ is 1.