Hello mr. Gosper, yes, very interesting, explicit formulas : this is what I need. I included your formula for the case k=2/7 on the 3rd version of my preprint here : http://arxiv.org/abs/1101.6066 (tell me if there is an error ?). thanks for all this, très respectueusement, Simon Plouffe ----------- Yes, that's a better way to write it. You could mention that all sums Sum[n^3/(E^(2*Pi*n/k) - 1), {n, Infinity}] for rational k seem to be 3 2 n 1 Pi A[k] Sum[----------------, {n, Infinity}] == -(---) + --------- (2 n Pi)/k 240 3 8 -1 + E Gamma[-] 4 for some algebraic number A[k]. E.g., for k=6, 3 n Sum[--------------, {n, Infinity}] == (n Pi)/3 -1 + E 1/4 3/4 2 1 (231 + 120 Sqrt[2] 3 + 140 Sqrt[3] + 60 Sqrt[2] 3 ) Pi -(---) + ------------------------------------------------------------ 240 3 8 320 Gamma[-] 4 so they are all readily found with PSLQ. For increasing k, they become very close to rationals, e.g., In[474]:= ContinuedFraction[%[[2]], 7] Out[477]= {5, 2, 1, 1, 9, 7896393272, 8} For k=13 (your 2.2), you could write Sum[n^3/(-1 + E^((2*n*Pi)/13)), {n, Infinity}] == -(1/240) + (1/Gamma[3/4]^8)*(Pi^2*Root[-2543222153227391999 + 1780150088904190080*#1 - 5828105673767424000*#1^2 - 1264509990338560000*#1^3 - 1050320673177600000*#1^4 - 48338514739200000*#1^5 + 1073741824000000*#1^6 & , 2]) I see by your remark following 2.7 that you already knew Sum[(1/(n * (E^(k * n * Pi) - 1))),{n, 1, Infinity}==Log[F(k)] ! Maybe you should state that formula explicitly. And you can redo 2.3 thru 2.7, since we know the values of all those F[k]. (We can even do a fair number with k=sqrt(rational). This should all go into FunctionExpand.) --rwg