This proposition states that given the functor $\underset{\longleftarrow}{\lim}^1 \colon Ab^{(\mathbb{N}, \geq)} \longrightarrow Ab$ and the short exact sequence $0 \to A_\bullet \to B_\bullet \to C_\bullet \to 0 \;\;\; \in Ab^{(\mathbb{N}, \geq)}$, the following long exact sequence is induced: $$0 \to \underset{\longleftarrow}{\lim}_n A_n \to \underset{\longleftarrow}{\lim}_n B_n \to \underset{\longleftarrow}{\lim}_n C_n \to \underset{\longleftarrow}{\lim}_n^1 A_n \to \underset{\longleftarrow}{\lim}_n^1 B_n \to \underset{\longleftarrow}{\lim}_n^1 C_n \to 0.$$
The proof of this proposition creates a short exact sequence of chain complexes. The long exact sequence is then the long exact sequence of cohomology groups where the $k^{th}$ cohomology groups are trivial for $k \geq 2$.
I now want to consider the long exact sequence induced by the functor $\underset{\longleftarrow}{\lim}^1 \colon Ab^{(\mathbb{N}, \geq)} \longrightarrow Ab$ and the exact sequence $0 \to A_\bullet \to B_\bullet \to C_\bullet \to D_\bullet \to 0 \;\;\; \in Ab^{(\mathbb{N}, \geq)}$.
I think that in this case, we will NOT have an induced sequence of the form: $$0 \to \underset{\longleftarrow}{\lim}_n A_n \to \underset{\longleftarrow}{\lim}_n B_n \to \underset{\longleftarrow}{\lim}_n C_n \to \underset{\longleftarrow}{\lim}_n D_n \to \underset{\longleftarrow}{\lim}_n^1 A_n \to \underset{\longleftarrow}{\lim}_n^1 B_n \to \underset{\longleftarrow}{\lim}_n^1 C_n \to \underset{\longleftarrow}{\lim}_n^1 D_n \to 0.$$
Is that correct? If so, What kind of sequence will be induced instead? (That's my question.) Will there be an induced spectral sequence? I admit that I am not yet familiar with spectral sequences, but the reason I suspect there will be one in the 4-term exact sequence case is because I read that while a long exact sequence of cohomology arises from a short exact sequence, a spectral sequence arises from a 4-term exact sequence.