Different methods to compute \sum\limits_{n=1}^\infty \frac{1}{n^2}

OK, here's my favorite. I thought of this after reading a proof from the book "Proofs from the book" by Aigner & Ziegler, but later I found more or less the same proof as mine in a paper published a few years earlier by Josef Hofbauer. On Robin's list, the proof most similar to this is number 9 (EDIT: ...which is actually the proof that I read in Aigner & Ziegler).
When 0<x<π/2 we have 0<sinx<x<tanx and thus

1tan2x<1x2<1sin2x.

Note that 1/tan2x=1/sin2x−1. Split the interval (0,π/2) into 2n equal parts, and sum the inequality over the (inner) "gridpoints" xk=(π/2)⋅(k/2n):

∑k=12n−11sin2xk−∑k=12n−11<∑k=12n−11x2k<∑k=12n−11sin2xk.

Denoting the sum on the right-hand side by Sn, we can write this as

Sn−(2n−1)<∑k=12n−1(2⋅2nπ)21k2<Sn.

Although Sn looks like a complicated sum, it can actually be computed fairly easily. To begin with,

1sin2x+1sin2(π2−x)=cos2x+sin2xcos2x⋅sin2x=4sin22x.

Therefore, if we pair up the terms in the sum Sn except the midpoint π/4 (take the point xk in the left half of the interval (0,π/2) together with the point π/2−xk in the right half) we get 4 times a sum of the same form, but taking twice as big steps so that we only sum over every other gridpoint; that is, over those gridpoints that correspond to splitting the interval into 2n−1 parts. And the midpoint π/4 contributes with 1/sin2(π/4)=2 to the sum. In short,

Sn=4Sn−1+2.

Since S1=2, the solution of this recurrence is

Sn=2(4n−1)3.

(For example like this: the particular (constant) solution (Sp)n=−2/3 plus the general solution to the homogeneous equation (Sh)n=A⋅4n, with the constant A determined by the initial condition S1=(Sp)1+(Sh)1=2.) We now have

2(4n−1)3−(2n−1)≤4n+1π2∑k=12n−11k2≤2(4n−1)3.

Multiply by π2/4n+1 and let n→∞. This squeezes the partial sums between two sequences both tending to π2/6. Voilà!