OFFSET
0,2
COMMENTS
An upper bound for the Lagrange remainder in the expansion of log((1+x)/(1-x)) for x=1/3, i.e., for log(2), is R(2*n) = (1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1), n >= 0.
The denominators are found in A096953.
log(2) = 2*Sum_{k>=1} ((1/3)^(2*k-1))/(2*k-1), from the Taylor series of log((1+x)/(1-x)) for x=1/3.
REFERENCES
M. Barner and F. Flohr, Analysis I, de Gruyter, 5. Auflage, 2000; p. 293.
LINKS
W. Lang, More comments.
FORMULA
a(n) = numerator(A(n)), where A(n) = (6/5)*(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1) = A096951(n)/((2*n+1)*6^(2*n)).
EXAMPLE
n=3: R(2*3)=(5/6)* a(3)/A096953(3) = (5/6)*463/326592 = 2315/1959552 = 0.001181..., therefore log(2) - 2*Sum_{k=1..3} ((1/3)^(2*k-1))/(2*k-1) < 0.001181... . In fact, the partial sum is 0.0001430654... .
CROSSREFS
KEYWORD
nonn,easy,frac
AUTHOR
Wolfdieter Lang, Jul 16 2004
STATUS
approved