Simple examples for the use of spectral sequences I'm looking for basic examples that show the usefulness of spectral sequences even in the simplest case of spectral sequence of a filtered complex. 
All I know are certain "extreme cases", where the spectral sequences collapses very early to yield the acyclicity of the given complex or some quasi-isomorphism to another easier complex (balancing tor, for example).
Is there an example of a useful filtration where one really computes something nontrivial also in the higher sheets?
The examples I have in mind come from topology. For example, the calculation of $H_{\ast}(\Omega{\mathbb S}^n;{\mathbb Z})$ is simply beautiful using the Serre spectral sequence, and one needs to pass to the $n$-th sheet until something happens. Another more difficult example would be the computation of the rational cohomology of $K({\mathbb Z},n)$ by induction on $n$ (depending on the parity of $n$, we get a polynomial algebra or an exterior algebra, if I remember correctly). 
Are there similar, but purely algebraic examples which could show the usefulness of spectral sequences to those seeing them the first time?
 A: This isn't exactly what you asked, but its a very simple example that (to me) demonstrates some of the necessity of the complexities of spectral sequences.  Consider the ring $R=\mathbb{C}[x,y]$, and consider the module $M=(Rx+Ry)\oplus R/x$.  Then the double dual spectral sequence converges to the original module:
$$ Ext^{-i}_R(Ext^j_R(M,R),R) \Rightarrow M $$
The second page of this spectral sequence has 


*

*$R$ in degree $(0,0)$

*$R/x$ in degree $(-1,1)$

*$R/(Rx+Ry)$ in degree $(-1,2)$

*A non-trivial knights-move map  (differential on the second page) from $(0,0)$ to $(-1,2)$ which is the natural quotient map.


The spectral sequence collapses on the third page, with $(Rx+Ry)$ in degree $(0,0)$ and $R/x$ in degree $(-1,1)$.  One shortcoming of this example is that you get the same module back, split apart into different components; rather than the associated graded of some interesting filtration.  If memory serves, there was a way to tinker with this example to give it that property too, but it escapes me at the moment.
A: Non-trivial spectral sequences occur when calculating the homotopy groups of tmf at 2 or 3: http://www.math.uiuc.edu/~rezk/512-spr2001-notes.pdf §16 ff., see also Tilman Bauer's article http://arxiv.org/abs/math/0311328
A: I can't recommend the following document by Tom Weston enough. It introduces spectral sequences rapidly and at a comfortable level of generality. It then applies the Hochschild-Serre sequence to group cohomology.
www.math.mcgill.ca/goren/SeminarOnCohomology/infres.pdf
A: The best example I can think of is the Lyndon-Hochschild-Serre spectral sequence in group cohomology. See for instance Chapter VII Section 6 of Brown's Cohomology of Groups.
The spectral sequence, for a group extension $1\to H\to G\to Q \to 1$ (I'm using Brown's notation) and for a G-module M, is of the form
$$E^2_{pq}=H_p(Q,H_q(H,M))\Rightarrow H_{p+q}(G,M).$$
Let's use this to calculate the third integral homology of the dihedral group D2n, for $n$ an odd integer. The group extension is $$1\to C_n\to D_{2n}\to C_2 \to 1,$$
and the corresponding Lyndon-Hochschild-Serre spectral sequence is 
$$E^2_{pq}=H_p(C_2,H_q(C_n,\mathbb{Z}))\Rightarrow H_{3}(D_{2n},\mathbb{Z}),$$
where p and q add up to 3. The integral homology of a cyclic group Cm is $\mathbb{Z}$ for $q=0$, vanishes for $q\in\mathbb{N}^\ast$ even, and is $\mathbb{Z}/m\mathbb{Z}$ for $q$ odd. Plug this information into the Lyndon-Hochschild-Serre spectral sequence, and you find
$$H_{3}(D_{2n},\mathbb{Z})=H_0(C_2,\mathbb{Z}/n\mathbb{Z})\oplus H_3(C_2,\mathbb{Z})\simeq \mathbb{Z}/2n\mathbb{Z}.$$
This is easy and elegant calculation in my opinion, and occurs in practice in knot theory ($H_{3}(D_{2n},\mathbb{Z})$ turns out to be isomorphic to the relative bordism group of Fox n-coloured knots).
A: This is not the most profound answer but it's something that came up last week when I was reading a paper. The author wanted to prove that a certain obstruction to a problem was zero, and the obstruction lived in an $H^2$ that looked a bit scary: it was $H^2(W,V)$ with $W$ a local Weil group and $V$ a finite-dimensional vector space over a field of characteristic zero (I'll tell you all you need to know about this Weil group in a sec, in case you don't know what one is; it's a topological group coming up in number theory). But then I realised the obstruction was really in the image of $H^2(W/C,V)$ with $C$ a compact open subgroup of $W$ (a finite index subgroup of inertia). 
But now I'm done because $W/C$ has a two-step filtration with a finite sub and a quotient isomorphic to $\mathbf{Z}$, and I know finite groups have no cohomology in char 0 in degrees 1 or more, and $\mathbf{Z}$ is the fundamental group of a 1-dimensional thing so it has no cohomology in degree 2 or more, and so by Hochschild-Serre, a calculation I can even do in my head in this example, there are no non-zero terms to build $H^2(W/C,V)$ from in $E_2$ and hence in $E_\infty$ so this group vanishes. I can do this calculation without even pulling out a piece of paper.
I'm sure if I were more "group-theoretic" I would have a much clearer picture about what was going on, but Hochschild-Serre just explains to me in a very concrete way how the cohomology of a group is built from the cohomology of its subs and quotients, and is definitely something I carry around in my "useful tools" bag.
A: This is not an answer in algebra, but I wrote the following without noticing that you had asked for examples in algebra.  So here it is.  
I like the Mayer-Vietoris spectral sequence for giving easy examples.  For simplicity, suppose that $K$ is a simplicial complex expressed as a union of $n$ subcomplexes $K_1,\ldots,K_n$.  You can make a long exact sequence of chain complexes, with the degree zero bit being the chain complex $C_*(K)$, the degree one bit being the direct sum $\bigoplus_i C_*(K_i)$, the degree two bit begin the direct sum $\bigoplus_{i<j}C_*(K_i\cap K_j)$, the degree three bit being $\bigoplus_{i<j<k}C_*(K_i\cap K_j\cap K_k)$, etc.  
For one of the two spectral sequences arising from this double chain complex, the $E^1$-page is all zero.  For the other one, the $E^1$-page is the homologies of $K$ and all of the intersections of the pieces.  If you only have two pieces, the final differential is $d^2$, which carries the same information as the connecting homomorphism in the standard Mayer-Vietoris long exact sequence.  
How can the spectral sequence be helpful?  I was once shown a `proof' that $H_m(K)=0$ for some $K$ made from $n$ pieces in this way, because $H_{m}(K_i)=0$ for all $i$ and $H_{m-1}$ of each intersection was 0.  Knowing about the spectral sequence makes it easy to see that there is a mistake in this argument: what you really need is that $H_{m-k}=0$ for each of the $k$-fold intersections for each $k$.  
For an example where a higher differential is non-zero, consider this spectral sequence for the reduced homology of the $(n-2)$-sphere, made as the boundary of an $(n-1)$-simplex.  This triangulation of the $(n-2)$-sphere consists of $n$ top dimensional simplices $K_1,\ldots,K_n$ and each intersection of some proper subset of these is a simplex, while the intersection of all $n$ of them is empty.  In the $E^1$-page of the non-trivial spectral sequence, every entry is zero except the zero column, which contains the reduced homology of $S^{n-2}$ and the $n$th column which contains the reduced homology of the empty set (i.e., $\mathbb{Z}$ in dimension $-1$ 
and 0 elsewhere).  The differential $d^n$ must therefore be an isomorphism, showing that the reduced homology of $S^{n-2}$ is zero except in degree $n-2$ where it is $\mathbb{Z}$.  
