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).

When0<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!

When

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