Different methods to compute ∞∑k=11k2\sum\limits_{k=1}^\infty \frac{1}{k^2} (Basel problem)

As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
ζ(2)=k=11k2=π26.
However, Euler was Euler and he gave other proofs.

I believe many of you know some nice proofs of this, can you please share it with us?

Answer

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):
2n1k=11sin2xk2n1k=11<2n1k=11x2k<2n1k=11sin2xk.
Denoting the sum on the right-hand side by Sn, we can write this as
Sn(2n1)<2n1k=1(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+1π22n1k=11k22(4n1)3.
Multiply by π2/4n+1 and let n. This squeezes the partial sums between two sequences both tending to π2/6. Voilà!

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