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

[Albert Chan]

(10-26-2021 02:12 AM)Gerson W. Barbosa Wrote:  Numerically. I computed the difference of the exact result and the partial sums for a few n and examined their inverses.
For example:

10 -> 10.507917
20 -> 20.504062
90 -> 90.500921

That gives the first term of the continued fraction, 1/(n + 1/2)

Interesting many CF correction formula involve the tag along +1/2

Quote:By doing the calculation with 34 digits, it was possible to get the next three terms:

1/((n + 1/2) + 1/(12*(n + 1/2) + 16/(5*(n + 1/2) + 81/(28*(n+1/2) + ... ))))

The numerators are obvious, but the coefficients of the denominators don't appear to follow a clear pattern ...

Impressive you can deduce denominator pattern with only 4 numbers: 1, 12, 5, 28, ...
Anther way to check units (dimensional analysis).

Numerator: 1, 16, 81, ... seems to involve unit of U^4
Denominator should have unit U^2, to have everything consistent.
But we only have unit U from (n+1/2), so coefficients somehow also have unit of U
To pull units off coefficients, we interpolate:

(1,1), (3,5) ⇒ y = 2*x-1 = 2*(x-0.5)
(2,12), (4,28) ⇒ y= 8*x-4 = 8*(x-0.5)

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) + ... ))))

Quote:

(10-25-2021 01:29 PM)Albert Chan Wrote:  This converge even faster, A cute sum of Ramanujan

zeta2(n) := (10 - sum(1/(k*(k+1))^3, k=1..n)) / 6

Is that ok? 50 terms of the above series give only 10 correct digits:

1.644934066(94481120)

Considering number of terms needed to sum 1/k^2, this is a huge improvement.
Maybe we can accelerate convergence with CF correction ?