PAGE 68 [...] The author is perplexed by the fact that terms in the right side of this equation blow up when x^j=y^(j+1) or x^(j+1)=y^j, while the left side does not. [...] This indicates that when such an offending j exists and is positive, the above expression [summing from j instead of 0] must be 0. NeilB finally got to the bottom of this. It's a strange, q-ish identity. At *least* for positive x, y, c, and m, if y = x^(m/(1 + m)) or x^(1 + 1/m), then Sum[(x^(2 + 4*k) + c*y^(2 + 4*k))* Product[(c*(x*y)^j*(x^(1 + 2*j) - y^(1 + 2*j))* (x^(2 + 2*j) - y^(2 + 2*j)))/ ((x^(2 + j) - y^(1 + j))*(x^(1 + j) - y^(2 + j))* (x^(2 + 2*j) + c*y^(2 + 2*j))*(x^(3 + 2*j) + c*y^(3 + 2*j))), {j, m, k - 1}], {k, m, Infinity}] == (x^(1 + 2*m) + c*y^(1 + 2*m))* (x*y)^(1 + m)*Sum[Product[ (c*(x*y)^(j + 1)*(x^(1 + 2*j) - y^(1 + 2*j))* (x^(2 + 2*j) - y^(2 + 2*j)))/ ((x^(1 + 2*j) + c*y^(1 + 2*j))*(x^(2 + j) - y^(1 + j))*(x^(1 + j) - y^(2 + j))* (x^(2 + 2*j) + c*y^(2 + 2*j))), {j, m, k - 1}], {k, m, Infinity}]; E.g., taking three terms with m=1/3, In[167]:= (List @@ %) /. y -> x^(1 + 1/m)/. {∞ -> 7/3, m -> 1/3, Sum -> Table} /. x -> N[E] /. c -> N[\[Pi]] Out[167]= {{1.93977*10^6, 4348.25, 0.465464}, {1.9439*10^6, 216.488, 0.00115377}} In[168]:= Total /@ % Out[168]= {1.94411*10^6, 1.94411*10^6} I should convert this to 2×2 matrix products with q and z. --rwg On Tue, Mar 4, 2014 at 8:13 PM, Bill Gosper <billgosper@gmail.com> wrote:
Also, by http://dspace.mit.edu/bitstream/handle/1721.1/6088/AIM-304.pdf
p25:
HarmonicNumber[p]==-Sum[(3*k + p)* Binomial[k - p - 1, k]/k/(k + p)/Binomial[2*k, k], {k, ∞}]
(Terminates after (integer) p), which Mma converts to two useless 3F2s
unless integer 2p.
--rwg
Ouch, there are way more typos in AIM 304 than I thought. Even
"some" instead of "sum". I wonder if MIT lost the errata.
Repairing one egregium, p66:
Sum[1/Fibonacci[n], {n, ∞}] == Sum[(I^((-1 + k)*k)* ((-1)^k*Fibonacci[2 + 2*k] + Fibonacci[3 + 4*k]))/ (Fibonacci[1 + 2*k]*Fibonacci[2 + 2*k]* Product[LucasL[1 + 2*j], {j, k}]),{k, 0, ∞}]
For some reason, this (rapidly convergent) rhs completely stymies Mma's numerics. It doesn't even yell False if you change LucasL[2j+1] to
LucasL[2j-1].
There shortly follows an acceleration of Sum 1/(x^n+cy^n) which
divides by zero if there is a nonnegative integer j s.t. x^j=y^(j+1)
or y^j=x^(j+1).
PUZZLE 1 (from back in my smarter days): rewrite Sum(1/(x^n-1)) (= Sum[DivisorSigma[0,n]/x^n])
so as to avoid the 1/0.
SPOILER Sum[1/(x^n - 1), {n, ∞}] == Sum[1/(x^(n/4) - (-1)^n) + (-1)^n/(x^(n/4) - 1), {n, ∞}]/2
PUZZLE 2: Prove it.
--rwg