Confused about Hypercohomology terminology and meaning check this: Given a sheaf complex $F^\bullet$, let's say I want to compute the hypercohomology of this complex, if we consider the bicomplex of sheaves
$C^\bullet(F^\bullet) = (C^p(F^q))\quad (p,q\in\mathbb{Z})$,
where $C^\bullet(F^q)$ is the Godement resolution of the sheaf $F^q$. The hypercohomology of $F^\bullet$ is the cohomology of the complex
$K^\bullet(X) = tot(C^\bullet(F^\bullet)(X))$.
If we use spectral sequences to compute the hypercohomology I have two spectral sequences, let's look at the first spectral sequence {$'E^{p,q}_r$}, this sequence converges to the final term $'E^{p,q}_\infty$ right?
This term is at the same time
$'E^{p,q}_\infty = Gr^p_C \: \mathbb{H}^{p + q}(K^\bullet(X)) = C^{p+1}(\mathbb{H}^{p + q}(K^\bullet(X)))/C^p(\mathbb{H}^{p + q}(K^\bullet(X)))$
right? This is commonly expressed as
$'E^{p,q}_2 = H^p(X,H^q(F^\bullet)) \Rightarrow \mathbb{H}^{p + q}(K^\bullet(X))$,
and the second spectral sequence {$''E^{p,q}_r$} also converges to this. Ok my questions now are:
1 - Some authors simply say that these spectral sequences converge to the hypercohomology $\mathbb{H}^{p + q}(K^\bullet(X))$, why do they say that if the spectral sequences clearly converge to $Gr^p_F \: \mathbb{H}^{p + q}(K^\bullet(X)) = C^{p+1}(\mathbb{H}^{p + q}(K^\bullet(X)))/C^p(\mathbb{H}^{p + q}(K^\bullet(X)))$? Or where am I wrong?
2 - Let's say I have succesfully computed ALL the terms in the two spectral sequences, what do I gain from obtaining $Gr^p_F \: \mathbb{H}^{p + q}(K^\bullet(X)) = C^{p+1}(\mathbb{H}^{p + q}(K^\bullet(X)))/C^p(\mathbb{H}^{p + q}(K^\bullet(X)))$? What is that telling me?, like what if $Gr^p_F \: \mathbb{H}^{p + q}(K^\bullet(X)) = 0$ for some $p$, and $q$? What can I get from knowing that $C^{p+1}(\mathbb{H}^{p + q}(K^\bullet(X)))/C^p(\mathbb{H}^{p + q}(K^\bullet(X)))$ is $0$? How can I use that to compute $\mathbb{H}^{p + q}(K^\bullet(X))$, which is actually what I'm looking for? I know it's dumb and am missing something but I can't see it, can anybody please help me understand this, thanks.
 A: *

*People generally say that a spectral sequence "converges" to something even though it generally only provides the associated graded object of the actual (filtered) thing you're after.

*Successfully running such a spectral sequence provides you with the information you want "up to extension problems".  The most basic example of an extension problem is if say you're after an $R$-module $M$, and I tell you that there's an exact sequence $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$, and I tell you what $M'$ and $M''$ are.  There is a beautiful theory of such extension problems: The set of isomorphism classes of such extensions $0 \rightarrow M' \rightarrow ? \rightarrow M'' \rightarrow 0$ form a group $\mbox{Ext}^1_R(M'',M')$, the extensions of $M''$ by $M'$.  The identity element is given by the trivial extension $M=M' \oplus M''$.  When this group vanishes, then you know that this is the only possibility!  For example, this group vanishes when $R$ is a field: then $R$-modules are just vector spaces, and these are completely classified by their dimension (and you can easily check that it must be that $\dim(M')+\dim(M'')=\dim(M)$).
You can read the wikipedia page on the "Ext functor" to find out how to compute it.  Perhaps it will help to know that this is the so-called derived functor of $\mbox{Hom}$.
