## 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/sin2x1. Split the interval (0,π/2) into 2n equal parts, and sum the inequality over the (inner) "gridpoints" xk=(π/2)(k/2n):
k=12n11sin2xkk=12n11<k=12n11x2k<k=12n11sin2xk.
Denoting the sum on the right-hand side by Sn, we can write this as
Sn(2n1)<k=12n1(22nπ)21k2<Sn.
Although Sn looks like a complicated sum, it can actually be computed fairly easily. To begin with,
1sin2x+1sin2(π2x)=cos2x+sin2xcos2xsin2x=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/2xk 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 2n1 parts. And the midpoint π/4 contributes with 1/sin2(π/4)=2 to the sum. In short,
Sn=4Sn1+2.
Since S1=2, the solution of this recurrence is
Sn=2(4n1)3.
(For example like this: the particular (constant) solution (Sp)n=2/3 plus the general solution to the homogeneous equation (Sh)n=A4n, with the constant A determined by the initial condition S1=(Sp)1+(Sh)1=2.) We now have
2(4n1)3(2n1)4n+1π2k=12n11k22(4n1)3.
Multiply by π2/4n+1 and let n. This squeezes the partial sums between two sequences both tending to π2/6. Voilà!