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A096952 Numerators of upper bounds for Lagrange remainder in Taylor's expansion of log((1+x)/(1-x)) for x=1/3, multiplied by 6/5. 3
1, 7, 11, 463, 4039, 35839, 320503, 575267, 25854247, 232557151, 298927153, 18830313487, 6778577311, 1525146340543, 13726182847159, 123535108753519, 1111813831298023, 2001263178349523, 90056808665990167, 810511140554958031 (list; graph; refs; listen; history; text; internal format)
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(((1/3)^(2*k-1))/(2*k-1),k=1..infty), 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

Table of n, a(n) for n=0..19.

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(((1/3)^(2*k-1))/(2*k-1), k=1..3) < 0.001181.. In fact, the partial sum is 0.0001430654..

CROSSREFS

Sequence in context: A201120 A164328 A301736 * A143602 A177999 A293220

Adjacent sequences:  A096949 A096950 A096951 * A096953 A096954 A096955

KEYWORD

nonn,easy,frac

AUTHOR

Wolfdieter Lang, Jul 16 2004

STATUS

approved

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Last modified June 18 14:52 EDT 2019. Contains 324213 sequences. (Running on oeis4.)