As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
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?
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
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):
Denoting the sum on the right-hand side by Sn, we can write this as
Although Sn looks like a complicated sum, it can actually be computed fairly easily. To begin with,
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,
Since S1=2, the solution of this recurrence is
(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
Multiply by π2/4n+1 and let n→∞. This squeezes the partial sums between two sequences both tending to π2/6. Voilà!