Is $\text{for}\ n\in \mathbb{N},\dfrac{u_{n+1}}{u_n}=q \iff \text{for}\ n\in \mathbb{N},u_{n+1}=q u_n$?

Question

In the correction of bac exams in my country , The Correction Committee ordered us to give the whole point for students who they showed that the sequence $u_n$ is geometric using $u_{n+1}=q u_n$ coming up to the value of $q$ in the same time they ordered us do not give the whole point for students who they follow this method :$ \text{for}\ n\in \mathbb{N},\dfrac{u_{n+1}}{u_n}=q $ for investigation of the value of $q$ they didn't accepted this method because they think that terms of geometric sequence might be nul by means probably the second term or third or .. would be nul and this disagree with logic because we can't divide over $0$ , In my opinion both method are true since the question is : show that sequence is geometric ? Now my question here is :

Question Are both methods true and useful at students to show such sequence is geometric or Is $\text{for}\ n\in \mathbb{N},\dfrac{u_{n+1}}{u_n}=q \iff \text{for}\ n\in \mathbb{N},u_{n+1}=q u_n$ ?

Answer 1

If $u_{n+1}/u_n = q$ for all positive integers $n$, then in particular $u_{n+1}/u_n$ must be defined, so none of the $u_n$ can be $0$. Thus $q$ can't be $0$.

On the other hand, if $u_{n+1} = q u_n$ for all $n$, you could have $q = 0$ with $u_0 = 1$ and $u_n = 0$ for all $n \ge 1$ (this is assuming your $\mathbb N$ includes $0$; if $\mathbb N$ starts with $1$, make it $u_1 = 1$ and $u_n = 0$ for $n \ge 2$).

Answer 2

If you want to prove that $$\forall n\in \Bbb N\;\;\frac{u_{n+1}}{u_n}=Cte=q$$

You should begin by showing that $$\forall n\in \Bbb N \;\; u_n\ne 0$$

This is not necessary to prove that there exist $ q\in \Bbb R $ such that $$\forall n\in \Bbb N \;\; u_{n+1}=qu_n$$

In the special case where $ q=0$, you can easily prove that $ u_{n+1}=0$ but you cannot prove that $$\frac{u_{n+1}}{u_n}=0$$