OFFSET
1,2
COMMENTS
For the partial sums of one half of this series, that is Sum_{k>=0} 1/((k+1)*(5*k+2)), with value 0.6613896265611944283..., see A294826(n)/A294827(n), for n >= 0. - Wolfdieter Lang, Nov 16 2017
REFERENCES
Max Koecher, Klassische elementare Analysis, Birkhäuser, Basel, Boston, 1987, Eulersche Reihen, pp. 189 - 193.
LINKS
L. Downey, B. W. Ong, and J. A. Sellers, Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers, Coll. Math. J. 39, no. 8, 2008, 391-394.
Society for Industrial and Applied Mathematics, Sums of Reciprocals of Polygonal Numbers and a Theorem of Gauss
Wikipedia, Heptagonal Number
FORMULA
Equals Sum_{n>=1} 2/(5n^2 - 3n).
((5/2)*log(5) - (2*phi-1)*(log(phi) - (Pi/5)*sqrt(7-4*phi)))/3, with the golden section phi := (1 + sqrt(5))/2. This is (5/10)*v_5(2) given from the Koecher reference on p. 192 as ((5/2)*log(5) - sqrt(5)*log((1+sqrt(5))/2) + (1/5)*Pi*sqrt(5*(5-2*sqrt(5))))/3. Compare this with the number given in the Mathematica program. - Wolfdieter Lang, Nov 16 2017
EXAMPLE
1.32277925312238885674944226131008401652280117371392437228545762688516221076....
MATHEMATICA
RealDigits[ Pi*Sqrt[25 - 10 Sqrt[5]]/15 + 2Log[5]/3 + (1 + Sqrt[5]) Log[ Sqrt[ 10 - 2 Sqrt[5]]/2]/3 + (1 - Sqrt[5]) Log[ Sqrt[ 10 + 2 Sqrt[5]]/2]/3, 10, 111][[1]] (* or *)
RealDigits[ Sum[2/(5 n^2 - 3 n), {n, 1, Infinity}], 10, 111][[1]]
PROG
(PARI) sumnumrat(2/n/(5*n-3), 1) \\ Charles R Greathouse IV, Feb 08 2023
CROSSREFS
KEYWORD
AUTHOR
Robert G. Wilson v, Jul 03 2014
STATUS
approved