Let $V=\left \{ (x_n)_{n\in\mathbb{Z}}\in\mathbb{R}^\mathbb{Z}\mid sup_{n\in\mathbb{Z}}\left | x_n \right | <\infty \right \}$
I want to show that $\left\Vert \cdot\right\Vert ^{'}=sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n+1\right|!}\}$ and $ \left\Vert \cdot\right\Vert =sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n\right|!}\}$ are not equivalent.
Which means that I need to find a series $v_n\in V$ which contradicts the assumption that $c\cdot sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n+1\right|!}\}\leq sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n\right|!}\}\leq C\cdot sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n+1\right|!}\}$ for some $c,C>0$ for all $v_n\in V$
My guess is that I need to somehow build $v_n$ such that $sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n+1\right|!}\} = 0$ and $sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n\right|!}\} > 0$ but I didnt quite manage to create such series..
Any Ideas?
Your general idea is correct, but $\sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n+1\right|!}\} = 0$ means $\forall n \in \mathbb Z: v_n=0$, so you have the zero vector of your vector space which has norm $0$ in any norm.
To disprove that
$$\sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n\right|!}\}\leq C\cdot \sup_{n\in\mathbb{Z}}\{\frac{\left|v_{n}\right|}{\left|n+1\right|!}\}$$
holds for any sequence and any given $C>0$, select an $N \in \mathbb Z, N > C$ and consider the vector
$$ v_n=\begin{cases} 1 & \text{ if } n=N\\ 0 & \text{ otherwise. } \\ \end{cases} $$
We have $\lVert (v_n)_{n \in \mathbb Z}\rVert=\frac1{N!}$ and $C\lVert (v_n)_{n \in \mathbb Z}\rVert^{'}=\frac{C}{(N+1)!} = \frac1{N!} \frac{C}{N+1} < \lVert (v_n)_{n \in \mathbb Z}\rVert,$
as $\frac{C}{N+1} < 1$ due the choice of $N$.
To disprove the other side of the inequality, the same kind of $(v_n)_{n \in \mathbb Z}$ works, you just need to select a negative $N < -\frac1{c}$.
OTOH, to disprove equivalence you only need to disprove one inequality. But since the index set is $\mathbb Z$, neither inequality holds. If we restrict the indix set to non-negative integers, then obviously $\lVert (v_n)_{n \in \mathbb N}\rVert^{'} \le \lVert (v_n)_{n \in \mathbb N}\rVert$ for all sequences.