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

[Albert Chan]

We can get the CF correction formula, using Euler–Maclaurin formula (see #9)

XCas> C(k) := bernoulli(k)/k!; // Euler-Maclaurin formula coefs
XCas> f, a, b := 1/x^2, N, inf; // N = n+1

XCas> corr := int(f,x,a,b)                       → 1/N
XCas> corr += preval(f,a,b) * (-1/2)       → 1/N+1/(N^2*2)
XCas> f := f' :; corr += preval(f,a,b) * C(k:=2)
XCas> f := f'':; corr += preval(f,a,b) * C(k+=2)

Run last line a few times, we have this corr, as polynomial of 1/N:

XCas> e2r(corr(N=1/x))

[-691/2730, 0, 5/66, 0, -1/30, 0, 1/42, 0, -1/30, 0, 1/6, 1/2, 1, 0]

Confirm numerically:

XCas> n:=5; sum(1./k^2, k=1..n) + corr(N=n+1), pi*pi/6.

(5, 1.64493406685, 1.64493406685)

Now, we are ready to convert corr into CF formula.
corr assumed N=n+1, but we wanted N=n+1/2, so we shift, and flip it.
Note, we replace N by N+1/2 in 1 step, instead of N by n+1 then n by N-1/2

XCas> [top,bot] := f2nd(1/corr(N=N+1/2)) :;
XCas> q:=quo(top,bot,N); [top,bot] := [bot,top-q*bot]:; → N

Run last line a few more times, we have the other quotients:

12*N, 5/16*N, 448/81*N, 729/4096*N, 180224/50625*N, 8125/65536*N

Convert simple CF to generalized CF, we have (note: now N = n+0.5)

corr = 1/(N+ 1/(12N + 16/(5N + 81/(28*N + 256/(9*N + 5^4/(44*N + 6^4/(13*N + ...