OFFSET
0,1
COMMENTS
If the contribution of the first term, -sin(1) = -A049469, is omitted, the constant becomes Sum_{n>=1} (sin(1/(2n)) - sin(1/(2n+1))) = 0.29067413667396703114243536419518058522...
LINKS
oldrinb, Evaluating the infinite series sum_n=1^infty [sin(1/2n)-sin(1/(2n+1))], math.stackexchange, Aug 22 2013
EXAMPLE
0.550796848133929475510066957...
MAPLE
M := 141 :
Digits := 120 :
s := sin(1/2/n)-sin(1/(2*n+1)) :
add(subs(n=i, s), i=1..M) :
pre := evalf(%) :
zetaM := proc(s, M)
local n ;
Zeta(s)-add(1/n^s, n=1..M) ;
evalf(%) ;
end proc:
for dd from 75 to 90 by 5 do
subs(n=1/eps, s) ;
taylor(%, eps=0, dd+1) ;
t := gfun[seriestolist](%, 'ogf') ;
add( op(j, t)*zetaM(j-1, M), j=3..nops(t)) ;
x := pre+% ;
print(x) ;
end do:
# now sum_{n>=1} (-1)^n*sin(1/n) = -0.5570986.
x-sin(1.0) ;
MATHEMATICA
digits = 105; NSum[(-1)^n*Sin[1/n], {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> digits+10] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 24 2014 *)
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
R. J. Mathar, Dec 08 2013
STATUS
approved