Evaluation of ζ(2) by the definition (sort of) [HP-42S & HP-71B]

[Albert Chan]

(10-26-2021 09:47 AM)Albert Chan Wrote:  It would be better if we had more coefficients to confirm the pattern.
But, this suggested "unit free" coefficients are alternating 2, 8, 2, 8 ...

1/(2*0.5*(n+0.5) + 1/(8*1.5*(n+0.5) + 16/(2*2.5*(n+0.5) + 81/(8*3.5*(n+0.5) + ... ))))

We can also sum odd terms: sum(1/(2k-1)^2, k=1..inf) = pi^2/8 = 3/4*ζ(2)

ζ(2) = 4/3 * sum(1/(2k-1)^2, k=1..inf)

If we sum only n terms, correction term ≈ ∫(1/(2x-1)^2, x=n+1/2 .. inf) = 1/(4n)

Doing numerically for the continued fraction form, we get a very similar form as above.

1/(4*n + 1/(2*1.5*n + 16/(8*2.5*n + 81/(2*3.5*n + 256/(8*4.5*n + ... )))))

Example, estimate ζ(2) with n=3

lua> n = 3
lua> 1 + 1/3^2 + 1/5^2
1.1511111111111112
lua> 1/(4*n + 1/(2*1.5*n + 16/(8*2.5*n + 81/(2*3.5*n))))
0.08258933029740148
lua> 4/3 * (1.1511111111111112 + 0.08258933029740148)
1.6449339218780168
lua> pi*pi/6
1.6449340668482264