[...] (42) and (43) raise the interesting question of evaluating
Product[((1/(12*n)) + 1)*((n - 1/2)*(E^(1/n) - 1))^n,{n, 1, Infinity}]
(i.e., prod(((n-1/2)*(%e^(1/n)-1))^n*(1+1/12/n),n,1,inf)
--"cwg"
Man, it's hard to get even a riesable value. .90058064316485891020 is
probably only good to 12 digits. --rwg
Foo, it's easy. Expand the prodand at oo (= Foo/F): In[259]:= Series[(1/(12*n) + 1)*((n - 1/2)*(E^(1/n) - 1))^n, {n, Infinity, 3}] Out[259]= (E^ SeriesData[n, Infinity, {Log[2], 0, -1/12, -1/24}, -1, 3, 1]* SeriesData[n, Infinity, {1, 1/12}, 0, 4, 1])/2^n That's not a series! In[260]:= Series[Normal[%], {n, Infinity, 3}] Out[260]= SeriesData[n, Infinity, {1, 0, -13/288, 1/5184}, 0, 4, 1] That's a series, and the n^-3 term is wrong! In[261]:= Series[Normal[Series[(1/(12*n) + 1)*((n - 1/2)*(E^(1/n) - 1))^n, {n, Infinity, 4}]], {n, Infinity, 3}] Out[261]= SeriesData[n, Infinity, {1, 0, -13/288, -409/25920}, 0, 4, 1] In[262]:= Normal[%] Out[262]= 1 - 409/(25920*n^3) - 13/(288*n^2) Construct two factors from this: In[263]:= %[[1 ;; 2]]*%[[{1, 3}]] Out[263]= (1 - 409/(25920*n^3))*(1 - 13/(288*n^2)) Dividing these out of the prodand In[264]:= Out[261]/% Out[264]= SeriesData[n, Infinity, {1}, 0, 4, 1] raises the convergence rate to n^-4. And we have a closed form for what we divided out: In[265]:= Product[(1 + 1/12/n)*((n - 1/2)*(E^(1/n) - 1))^n, {n, Infinity}] == Product[(1 + 1/12/n)*(((n - 1/2)*(E^(1/n) - 1))^n/%%), {n, Infinity}]* Product[%%, {n, Infinity}] Out[265]= Product[(1 + 1/(12*n))*((-1 + E^(1/n))*(-(1/2) + n))^n, {n, Infinity}] == (12*Sqrt[2/13]* Product[((1 + 1/(12*n))*((-1 + E^(1/n))*(-(1/2) + n))^n)/ ((1 - 409/(25920*n^3))*(1 - 13/(288*n^2))), {n, Infinity}]*Sin[(1/12)*Sqrt[13/2]*Pi])/ (Pi*Gamma[1 - (1/12)*(409/15)^(1/3)]* Gamma[1 + (1/24)*(409/15)^(1/3) - (I*(409/5)^(1/3))/(8*3^(5/6))]* Gamma[1 + (1/24)*(409/15)^(1/3) + (I*(409/5)^(1/3))/(8*3^(5/6))]) But o(n^-4) isn't quite enough for Mma to get 22 digits, say. Dividing out binomials for the next few terms got Julian 0.900580643164907990153824318444289111926734003669821962 + 0.*10^-55 I fairly quickly. Unfortunately, this only agrees with the results below to about 30D. Alternatively, you can (theoretically) eliminate arbitrarily many terms just by fooling with the first order term! E.g., with parameters Out[219]= {e -> (45240 - 2153*Sqrt[3770])/221676, f -> (45240 + 2153*Sqrt[3770])/221676, b -> (1/340)*(-110 + Sqrt[3770]), a -> (1/340)*(-110 - Sqrt[3770])} In[220]:= FullSimplify[ Series[Normal[ Series[(1 + a/n)^e*(1 + b/n)^f*((n - 1/2)*(E^(1/n) - 1))^n /. %, {n, \[Infinity], 6}]], {n, \[Infinity], 5}]] Out[220]= 16633 SeriesData[n, Infinity, {1, 0, 0, 0, 0, -(---------)}, 0, 6, 1] 308448000 is O(n^-5). We have Product[(1 + 1/12/n)*((n - 1/2)*(E^(1/n) - 1))^n, {n, \[Infinity]}]-> Product[(1 + a/n)^e*(1 + b/n)^f*((n - 1/2)*(E^(1/n) - 1))^n, {n, \[Infinity]}]/ Product[(1 + a/n)^e*(1 + b/n)^f/(1 + 1/12/n), {n, \[Infinity]}] Plugging in a,b,f,g, the rational function product becomes a pile of Gammas, and In[268]:= N[%[[2]], 33] Out[268]= 0.900580643164907990153824318444627 took just a few seconds. Here are two further runs with timings: Timing[N[%%[[2]],44]] {19.5682, 0.9005806431649079901538243184443488363691376176930701107} Timing[N[%%%[[2]],69]] {20.0368, 0.900580643164907990153824318444348836369137617068042914805385390880213} NProduct seems to be suffering from the cancellation in e^(1/n)-1. For integer k>1, Julian worked out the formula for Product[1-a/n^k,{n, 1, Infinity}] This would be especially interesting if we could find among the profusion of ak, ek, and fk satisfying Product[1-ak/n^ek)^fk,{k,oo}] = ((n - 1/2)*(E^(1/n) - 1))^n any which are predicable functions of k. --rwg