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Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4*k+1).
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%I #58 Oct 15 2024 04:23:41

%S 8,6,6,9,7,2,9,8,7,3,3,9,9,1,1,0,3,7,5,7,3,9,9,5,1,6,3,8,8,2,8,7,0,7,

%T 1,3,6,5,2,1,7,5,3,6,7,3,4,5,2,4,4,9,0,4,3,3,5,0,3,1,8,3,8,9,1,7,6,3,

%U 9,3,5,1,4,1,0,9,4,1,3,2,9,0,5,5,7,5,0,4,0,3,4,6,3,4,0,8,9,6,8,7,0,5,2,1,8

%N Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4*k+1).

%D Jolley, Summation of Series, Dover (1961) eq 82 page 16.

%D Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.17

%H Ivan Panchenko, <a href="/A181048/b181048.txt">Table of n, a(n) for n = 0..1000</a>

%H J. M. Borwein, P. B. Borwein, and K. Dilcher, <a href="http://www.jstor.org/stable/2324715">Pi, Euler numbers and asymptotic expansions</a>, Amer. Math. Monthly, 96 (1989), 681-687.

%H Eric W. Weisstein, <a href="https://mathworld.wolfram.com/EulersSeriesTransformation.html">Euler's Series Transformation</a>.

%H Herbert S. Wilf, <a href="https://doi.org/10.46298/dmtcs.265">Accelerated series for universal constants, by the WZ method</a>, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).

%F Equals (A093954 + A091648/sqrt(2))/2.

%F Integral_{x = 0..1} 1/(1+x^4) = Sum_{k >= 0} (-1)^k/(4*k+1) = (log(1+sqrt(2)) + Pi/2)/(2*sqrt(2)).

%F 1 - 1/5 + 1/9 - 1/13 + 1/17 - ... = (Pi*sqrt(2))/8 + (sqrt(2)*log(1 + sqrt(2)))/4 = (Pi + 2*log(1 + sqrt(2)))/(4 sqrt(2)). The first two are the formulas as given in Spiegel et al., the third is how Mathematica rewrites the infinite sum. - _Alonso del Arte_, Aug 11 2011

%F Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) ) ~ 1/N + 1/N^2 - 3/N^3 - 11/N^4 + 57/N^5 + 361/N^6 - - ..., where the sequence of coefficients [1, 1, -3, -11, 57, 361, ...] is A188458. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 1/4. An example is given below. Cf. A181049. - _Peter Bala_, Sep 23 2016

%F Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 1) = Sum_{n >= 0} 2^(n-1)*n!/A007696(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k + 1)). - _Peter Bala_, Dec 01 2021

%F From _Peter Bala_, Oct 23 2023: (Start)

%F The slowly converging series representation Sum_{n >= 0} (-1)^n/(4*n + 1) for the constant can be accelerated to give the following faster converging series:

%F 1/2 + 2*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5));

%F 7/10 + 8*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9));

%F 71/90 + 48*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13));

%F 971/1170 + 384*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13)*(4*n + 17)).

%F These results may be easily verified by taking the partial fraction expansions of the summands. The general result appears to be that for r >= 0, the constant equals

%F C(r) + (2^r)*r!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 5)*...*(4*n + 4*r + 1)), where C(r) is the rational number Sum_{k = 0..r-1} 2^(k-1)*k!/(1*5*9*...*(4*k + 1)). [added 19 Feb 2024: the general result can be proved by the WZ method as described in Wilf.]

%F In the limit as r -> oo we find that the constant equals Sum_{k >= 0} 2^(k-1)*k!/(Product_{i = 0..k} 4*i + 1) as noted above. (End)

%F From _Peter Bala_, Mar 03 2024: (Start)

%F Continued fraction: 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler.

%F Equals hypergeom([1/4, 1], [5/4], -1).

%F Gauss's continued fraction: 1/(1 + 1^2/(5 + 4^2/(9 + 5^2/(13 + 8^2/(17 + 9^2/(21 + 12^2/(25 + 13^2/(29 + 16^2/(33 + 17^2/(37 + ... )))))))))). (End)

%e 0.86697298733991103757399516388287071365217536734524490433....

%e At N = 100000 the truncated series Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) ) = 1.7339(3)5974(5)7982(5)075(25)79(846)27(404)7... to 32 digits The bracketed numbers show where this decimal expansion differs from that of 2*A181048. The numbers 1, 1, -3, -11, 57, 361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) ) = 1.7339(4)5974(6)7982(2)075(14)79(903)27(765)7.... - _Peter Bala_, Sep 23 2016

%t RealDigits[(Pi Sqrt[2])/8 + (Sqrt[2] Log[1 + Sqrt[2]])/4, 10, 100][[1]] (* _Alonso del Arte_, Aug 11 2011 *)

%o (PARI) (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) \\ _G. C. Greubel_, Jul 05 2017

%o (PARI) (asinh(1)+Pi/2)/sqrt(8) \\ _Charles R Greathouse IV_, Jul 06 2017

%Y Cf. A001586, A003881, A024396, A091648, A093954, A113476, A181049, A181122, A188458, A193534, A262246.

%K nonn,cons

%O 0,1

%A _Jonathan D. B. Hodgson_, Oct 01 2010, Oct 06 2010