By LatticeReduce, it appears that these are all of the form -1/240 + <algebraic>*Gamma[1/4]^8/Pi^6, e.g. for k=6, Sum[n^3/(E^(2*Pi*n/k)-1),{n,Infinity}== -1/240 + ((231 + 120 Sqrt[2] 3^(1/4) + 140 Sqrt[3] + 60 Sqrt[2] 3^(3/4)) Gamma[1/4]^8)/(5120 Pi^6) and Sum[n^3/(E^(2*Pi*n/7)-1),{n,Infinity}== 1/4 3/4 1 8 (301 + 210 Sqrt[2] 7 + 120 Sqrt[7] + 90 Sqrt[2] 7 ) Gamma[-] 1 4 -(---) + ------------------------------------------------------------------ 240 6 5120 Pi In[135]:= N[%, 33] Out[135]= 10.0000000000000001901617678886627 For k=13, the algebraic is (1/402653184000)(188822323200 + 91226112000 Sqrt[13] + 3145728000 (2128581 + 50505 Sqrt[3] + 610709 Sqrt[13] + 22176 Sqrt[39])^(1/3) + (66260343848333767461568512000000000 + 19010687557237458566381568000000000 Sqrt[13] - 31128880624384868352000000000 (50505 Sqrt[3] + 22176 Sqrt[39]))^(1/3)), which I suspect simplifies. k=163 might be, to quote Prof Watson, "Too cumbrous to be of any importance". --rwg Simon says:-> Hello everybody, these days I am working on exponential sums and I have found something of interest, like infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 13 = 119.00000000000000000000000000000009593745851025547335588584913... the precision is 31 digits. Another one is infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 7 = 10.0000000000000001901617678886626755843593058554453334802548978434061099438\ The precision here is 15 digits. For an argument of 2*Pi*n/163, the precision is 435 digits! I don't know why I took 163... (of course). in general, these sums will be near an integer if the argument is 2*Pi*n/k and k is NOT a multiple of 2,3 or 5, strange isn't ? These are the simplest I could find, if the exponent is of the form 4m-1 then the sum is often an integer with the same conditions. This, I believe extends a little bit the known formula of Ramanujan/Berndt/etc. A good question is : does someone has a simple explanation of this ? I don't. I am preparing a paper on these results. Because, I do have another one which is EXACT, namely for pi: In general these sums with a fractional exponent are very close to integers or pi, I have a couple of series for 1/pi, 1/pi^2, etc. here is the EXACT formula with fractional arguments mixed with integer exponents. /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | 10 | ) -----------------| - 40 | ) -------------------| | / n (exp(Pi n) - 1)| | / n (exp(2 Pi n) - 1)| | ----- | | ----- | \ n = 1 / \ n = 1 / /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 10 | ) -------------------| - 10 | ) -----------------| | / n (exp(4 Pi n) - 1)| | / / Pi n \| | ----- | | ----- n |exp(----) - 1|| \ n = 1 / \ n = 1 \ 5 // /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 40 | ) -------------------| - 10 | ) -------------------| | / / 2 Pi n \| | / / 4 Pi n \| | ----- n |exp(------) - 1|| | ----- n |exp(------) - 1|| \ n = 1 \ 5 // \ n = 1 \ 5 //
evalf(%); 3.14159265358979323846264338327950288419716939937510582097494459230781640628\ 62089986280348253421170679821480865132823066470938442... well, I verified up to 1000 digits and it holds.
This is trivial, ? I do not see how. If someone has a piece of information on why this exist, I would be glad to ear from it, references, known results, etc. Me, I never saw these kind of formulas before, have a good day, Simon Plouffe