Let the following recurrence relation be given:
$b_0 = 1$
$b_1 = -4$
$b_k = -4 \frac{n}{k} \left( b_{k-1} + b_{k-2} \right)$ with $ k \leq n$ and $k,n \in \mathbb{N}$
$n$ is a constant.
Question: How can I prove that $b_k \neq 0$ ?
So far, I am able to prove that $b_k \neq 0$ if $b_k$ would be defined without the fraction as:
$b_k = -4 \left( b_{k-1} + b_{k-2} \right)$
But from there on I am unable to describe the effect of the fraction $\frac{n}{k} \geq 1$.
Indeed we have $|b_k|≥2|b_{k-1}|$ in the range.
This is clear for small $k$, and we proceed by induction. Suppose that we have shown that $|b_{k-1}|≥2|b_{k-2}|$.
Then we have $$|b_k|≥4|b_{k-1}+b_{k-2}|$$
(note: this is where we use $n≥k$)
Now, since $|b_{k-2}|≤\frac 12|b_{k-1}|$ we see that $$|b_{k-1}+b_{k-2}|≥|b_{k-1}|-|b_{k-2}|≥\frac 12|b_{k-1}|$$ and we are done.