On Sun, Jan 23, 2011 at 2:30 PM, Bill Gosper <billgosper@gmail.com> wrote:
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.
By Rich's ancient number recognizer, sum(1/n/(e^(pi k n)-1)) all seem to be 3/4 ln pi -ln gamma(1/4) - k pi/24 + ln <algebraic>, e.g.
SUM(1/(N*(%E^(%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+7*LOG(2)/8-%PI/24 SUM(1/(N*(%E^(2*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+LOG(2)-%PI/12 SUM(1/(N*(%E^(3*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+7*LOG(SQRT(3)+1)/12+LOG(2-SQRT(2)*3^(1/4))/4+3*LOG(3)/8+11*LOG(2)/24-%PI/8 SUM(1/(N*(%E^(%PI*N/3)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+LOG(SQRT(3)+1)/12-LOG(2-SQRT(2)*3^(1/4))/4-LOG(3)/8+23*LOG(2)/24-%PI/72 SUM(1/(N*(%E^(4*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+11*LOG(2)/8-%PI/6 SUM(1/(N*(%E^(5*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4-LOG(5^(3/4)-5^(1/4)+2)/2+LOG(5)/2+11*LOG(2)/8-5*%PI/24 SUM(1/(N*(%E^(6*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+LOG(SQRT(3)+1)/6+3*LOG(3)/8+11*LOG(2)/12-%PI/4 which should imply your pi conjecture. --rwg
/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