A113476 -id:A113476 - cp's OEIS frontend

A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).

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, 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, 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

Offset: 0

Jonathan D. B. Hodgson, Oct 01 2010, Oct 06 2010

			0.86697298733991103757399516388287071365217536734524490433....
At N = 100000 the truncated series 2*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

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

Ivan Panchenko, Table of n, a(n) for n = 0..1000
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric W. Weisstein, Euler's Series Transformation.
Herbert S. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).

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

RealDigits[(Pi Sqrt[2])/8 + (Sqrt[2] Log[1 + Sqrt[2]])/4, 10, 100][[1]] (* Alonso del Arte, Aug 11 2011 *)

(log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) \\ G. C. Greubel, Jul 05 2017

(asinh(1)+Pi/2)/sqrt(8) \\ Charles R Greathouse IV, Jul 06 2017

Equals (A093954 + A091648/sqrt(2))/2.
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)).
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
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
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
From Peter Bala, Oct 23 2023: (Start)
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:
1/2 + 2*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5));
7/10 + 8*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9));
71/90 + 48*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13));
971/1170 + 384*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13)*(4*n + 17)).
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
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.]
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)
From Peter Bala, Mar 03 2024: (Start)
Continued fraction: 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler.
Equals hypergeom([1/4, 1], [5/4], -1).
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)

A181049 Decimal expansion of (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+3).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 01 2010, Oct 05 2010

Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*((Pi/2 - log(1 + sqrt(2)))/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3)) ~ 1/N - 1/N^2 - 3/N^3 + 11/N^4 + 57/N^5 - ..., where the sequence of coefficients [1, -1, -3, 11, 57, ...] is A212435. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 3/4. An example is given below. Cf. A181048. - Peter Bala, Sep 23 2016

			0.2437477471996805241799750836323027110...
From _Peter Bala_, Sep 23 2016: (Start)
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3) = 0.4874(8)5494(4)9936(4)048(24)99(444)67(625)6... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181049. 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*( Pi/2 - log(1+sqrt(2)))/(2*sqrt(2) ) = 0.4874(9)5494(3)9936(1)048(35)99(501)67(264)6.... (End)

Gheorghe Coserea, Table of n, a(n) for n = 0..2015
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric W. Weisstein, Euler's Series Transformation.

Cf. A113476, A001586, A093954, A181048, A212435, A247719.

C := ComplexField(); [(Pi(C)/2 - Log(1+Sqrt(2)))/(2*Sqrt(2))]; // G. C. Greubel, Nov 28 2017

Mathematica

First@ RealDigits[N[(Pi/2 - Log[1 + Sqrt@ 2])/(2 Sqrt@ 2), 105]] (* Michael De Vlieger, Oct 07 2015 *)

PARI

default(realprecision, 106); eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+3)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015

PARI

(Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) \\ G. C. Greubel, Nov 28 2017

Formula

Equals Integral_{x=0..1} (x^2 dx)/(1+x^4).

Equals (1/2) * Integral_{x = 0..Pi/4} sqrt(tan(x)) dx. Cf. A247719. - Peter Bala, Sep 23 2016

Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 3) = Sum_{n >= 0} 2^(n-1)*n!/A008545(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k+3)). - Peter Bala, Dec 01 2021

From Peter Bala, Mar 03 2024: (Start)

Continued fraction: 1/(3 + 3^2/(4 + 7^2/(4 + 11^2/(4 + 15^2/(4 + ... ))))) due to Euler.

Equals (1/3)*hypergeom([3/4, 1], [7/4], -1).

Gauss's continued fraction: 1/(3 + 3^2/(7 + 4^2/(11 + 7^2/(15 + 8^2/(19 + 11^2/(23 + 12^2/(27 + 15^2/(31 + 16^2/(35 + 19^2/(39 + ... )))))))))). (End)

Equals Integral_{x=1..oo, y=1..oo} 1/(x^4 + y^4) dx. - Vaclav Kotesovec, Jun 13 2024

A193534 Decimal expansion of (1/3) * (Pi/sqrt(3) - log(2)).

Original entry on oeis.org

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

Offset: 0

Alonso del Arte, Jul 29 2011

The formulas for this number and the constant in A113476 are exactly the same except for one small, crucial detail: the infinite sum has a denominator of 3i + 2 rather than 3i + 1, while in the closed form, log(2)/3 is subtracted from rather than added to (Pi * sqrt(3))/9.

Understandably, the typesetter for Spiegel et al. (2009) set the closed formula for this number incorrectly (as being the same as for A113476, compare equation 21.16 on the same page of that book).

			0.373550727891424180392282045394659721402855371244161773816401641964909853052219...

L. B. W. Jolley, Summation of Series, Dover, 1961, eq. (80), page 16.
J. Rivaud, Analyse, Séries, équations différentielles, Mathématiques supérieures et spéciales, Premier cycle universitaire, Vuibert, 1981, Exercice 3, p. 132.
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 21.18.

Gheorghe Coserea, Table of n, a(n) for n = 0..2015
Eric W. Weisstein, Euler's Series Transformation.

Cf. A073010, A193535, A024396, A113476, A196548, A258969.

evalf((Psi(5/6)-Psi(1/3))/6, 120); # Vaclav Kotesovec, Jun 16 2015

RealDigits[(Pi Sqrt[3])/9 - (Log[2]/3), 10, 100][[1]]

(Pi/sqrt(3)-log(2))/3 \\ Charles R Greathouse IV, Jul 29 2011

default(realprecision, 98);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(3*n+2)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015

Equals Sum_{k >= 0} (-1)^k/(3k + 2) = 1/2 - 1/5 + 1/8 - 1/11 + 1/14 - 1/17 + ... (see A016789).
From Peter Bala, Feb 20 2015: (Start)
Equals (1/2) * Integral_{x = 0..1} 1/(1 + x^(3/2)) dx.
Generalized continued fraction: 1/(2 + 2^2/(3 + 5^2/(3 + 8^2/(3 + 11^2/(3 + ... ))))) due to Euler. For a sketch proof see A024396. (End)
Equals (Psi(5/6)-Psi(1/3))/6. - Vaclav Kotesovec, Jun 16 2015
Equals Integral_{x = 1..infinity} 1/(1 + x^3) dx. - Robert FERREOL, Dec 23 2016
Equals (1/2)*Sum_{n >= 0} n!*(3/2)^n/(Product_{k = 0..n} 3*k + 2) = (1/2)*Sum_{n >= 0} n!*(3/2)^n/A008544(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(3*k + 2)). - Peter Bala, Dec 01 2021
From Bernard Schott, Jan 28 2022: (Start)
Equals Integral_{x = 0..1} x/(1+ x^3) dx (see Rivaud reference).
Equals 3 * A196548. (End)
From Peter Bala, Mar 03 2024: (Start)
Equals (1/2)*hypergeom([2/3, 1], [5/3], -1).
Gauss's continued fraction: 1/(2 + 2^2/(5 + 3^2/(8 + 5^2/(11 + 6^2/(14 + 8^2/(17 + 9^2/(20 + 11^2/(23 + 12^2/(26 + ... ))))))))). (End)

A193535 Decimal expansion of log(2)/3.

Original entry on oeis.org

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

Offset: 0

Alonso del Arte, Jul 29 2011

This number is involved as an addend or subtrahend in the closed forms of certain series of reciprocals of integers (see for example A113476).

			0.231049060186648...

L. B. W. Jolley, Summation of Series, Dover (1961).
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, equations 21.16 and 21.18.

Index entries for transcendental numbers

Cf. A052502, A113476.

Mathematica
```
RealDigits[(Log[2]/3), 10, 100][[1]]
```

log(2)/3 \\ Charles R Greathouse IV, Jul 29 2011

Equals lim_{n->oo} [Sum_{i = 1..n} i^2/(n^3 + i^3)]. [Jolley eq 292, p.52]
Equals Sum_{n>=1} (-1)^(n-1)/(n*2^n*binomial(2*n, n)). - Arkadiusz Wesolowski, Jan 20 2013
From Amiram Eldar, Aug 05 2020: (Start)
Equals Integral_{x=1..oo} 1/(x^4 + x) dx.
Equals Integral_{x=0..oo} 1/(exp(2*x) + 3) dx. (End)
From Peter Bala, Feb 27 2024: (Start)
Equals (1/2)*Sum_{k >= 0} (-1)^k/((3*k + 1)*(3*k + 2)) = (1/2)*(1/(2 + (1*2)^2/(18 + (4*5)^2/(2*18 + (7*8)^2/(3*18 + (10*11)^2/(4*18 +  ... )))))) (continued fraction). See A052502.
Equals 7/32 + (3/2)*Sum_{k >= 0} (-1)^k/((3*k + 1)*(3*k + 2)*(3*k + 3)*(3*k + 4)*(3*k + 5)). (End)

A024217 a(n) = ( Product {k = 1..n} 3k - 2 ) ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 2) ).

Original entry on oeis.org

1, 3, 25, 222, 3166, 47016, 951544, 19827408, 520029520, 13952218560, 449559799360, 14756761434240, 563961412362880, 21893890640563200, 968019931702297600, 43385863589508249600, 2178487766250586470400, 110704921777161066700800, 6222745685273069016064000

Offset: 1

Clark Kimberling

Original name was: s(1)*s(2)*...*s(n)(1/s(1) - 1/s(2) + ... + c/s(n)), where c=(-1)^(n+1) and s(k) = 3k-2 for k = 1,2,3,...

L. Euler, De fractionibus continuis observationes, The Euler Archive, Index Number 123, Section 5

Cf. A007559, A024199, A024383, A024396, A113476.

I:=[1,3]; [n le 2 select I[n] else 3*Self(n-1)+(3*n-5)^2*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015

a[1] := 1: a[2] := 3: for n from 3 to 20 do a[n] := 3*a[n-1]+(3*n-5)^2*a[n-2] end do: seq(a[n], n = 1 .. 20); # Peter Bala, Feb 20 2015

Table[Product[3*k-2,{k,1,n}] * Sum[(-1)^(k+1)/(3*k-2),{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Feb 21 2015 *)

From Peter Bala, Feb 20 2015: (Start)
a(n) = A007559(n) * Sum {k = 1..n} (-1)^(k+1)/(3*k - 2).
Recurrence: a(n+1) = 3*a(n) + (3*n - 2)^2*a(n-1) with a(1) = 1 and a(2) = 3.
The triple factorial numbers A007559 also satisfy this second-order recurrence equation. This leads to the continued fraction representation a(n)/A007559(n) = 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + ... + (3*n - 2)^2/(3 ))))).
Taking the limit as n -> infinity gives the generalized continued fraction: Sum {k >= 1} (-1)^(k+1)/(3*k - 2) = 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + 10^2/(3 +  ... ))))) due to Euler.
The alternating sum has the value 1/3*( log(2) + Pi/sqrt(3) ) = A113476. Cf. A024396. (End)
a(n) ~ sqrt(2*Pi) * (sqrt(3)*Pi + 3*log(2)) * 3^(n-2) * n^(n-1/6) / (GAMMA(1/3) * exp(n)). - Vaclav Kotesovec, Feb 21 2015

New name from Peter Bala, Feb 20 2015

a(18)-a(19) from Vincenzo Librandi, Feb 21 2015

A181122 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+1).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 05 2010

			0.88831357265178863804075522702037934627811083077545817120597...

Gheorghe Coserea, Table of n, a(n) for n = 0..1000
H. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science 3(4) (1999), 189-192.

Cf. A002162, A003881, A024396, A113476, A181048, A181049, A181122, A193534, A262246.

(int(1/(1+x^5),x=0..1));
evalf(LerchPhi(-1,1,1/5)/5) ; # R. J. Mathar, Oct 16 2011

(Sqrt[8 + 8/Sqrt[5]]*Pi + 2*Sqrt[5]*ArcCoth[3/Sqrt[5]] + Log[16])/20 // RealDigits[#, 10, 105]& // First (* Jean-François Alcover, Feb 13 2013 *)

default(realprecision, 106);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(5*n+1)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015

Sum_{k>=0} (-1)^k/(5k+1) = Integral_{x=0..1}dx/(1+x^5) = (1/10)*sqrt(10-2*sqrt(5))*arctan((3/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*sqrt(10-2*sqrt(5))*arctan(-(1/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*sqrt(10-2*sqrt(5))*sqrt(5)*arctan(-(1/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*log(2)*sqrt(5) + (1/5)*log(2) - (1/20)*log(7-3*sqrt(5))*sqrt(5).
Equals Pi*sqrt(phi)/5^(5/4) + log(phi)/sqrt(5) + log(2)/5, where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Nov 01 2015
From Peter Bala, Feb 19 2024: (Start)
Equals (1/2)*Sum_{n >= 0} n!*(5/2)^n/(Product_{k = 0..n} 5*k + 1) = (1/2)*Sum_{n >= 0} n!*(5/2)^n/A008548(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(5*k + 1)).
Continued fraction: 1/(1 + 1^2/(5 + 6^2/(5 + 11^2/(5 + ... + (5*n + 1)^2/(5 + ... ))))).
The slowly converging series representation Sum_{n >= 0} (-1)^n/(5*n + 1) for the constant can be accelerated to give the following faster converging series:
1/2 + (5/2)*Sum_{n >= 0} (-1)^n/((5*n + 1)(5*n + 6)) and
17/24 + (25/2)*Sum_{n >= 0} (-1)^n/((5*n + 1)(5*n + 6)*(5*n + 11)).
These two series are the cases r = 1 and r = 2 of the general result: for r >= 0, the constant equals
C(r) + ((5/2)^r)*r!*Sum_{n >= 0} (-1)^n/((5*n + 1)*(5*n + 6)*...*(5*n + 5*r + 1)), where C(r) is the rational number (1/2)*Sum_{k = 0..r-1} (5/2)^k*k!/(1*6*11*...*(5*k + 1)). The general result can be proved by the WZ method as described in Wilf. (End)
From Peter Bala, Mar 03 2024: (Start)
Equals hypergeom([1/5, 1], [6/5], -1).
Gauss's continued fraction: 1/(1 + 1^2/(6 + 5^2/(11 + 6^2/(16 + 10^2/(21 + 11^2/(26 + 15^2/(31 + 16^2/(36 + 20^2/(41 + 21^2/(46 + ... )))))))))). (End)

A262246 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+2).

Original entry on oeis.org

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

Offset: 0

Gheorghe Coserea, Oct 06 2015

			0.4069016342...

Gheorghe Coserea, Table of n, a(n) for n = 0..2015
H. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science 3(4) (1999), 189-192.

Cf. A003881, A024396, A113476, A181048, A181049, A181122, A193534.

N[(1/5)*((Sqrt[5]-1)*Log[2] + Sqrt[5]*Log[Sin[3*Pi/10]] + (Pi/2)*Sec[Pi/10]), 100] (* G. C. Greubel, Oct 07 2015 *) (* fixed by Vaclav Kotesovec, Dec 11 2017 *)

default(realprecision, 87);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(5*n+2)))), "3..-2"))

Sum_{n>=0} (-1)^n/(5n+2) = Integral_{x=0..1} x/(1+x^5)dx.
From G. C. Greubel, Oct 07 2015: (Start)
Sum_{n>=0} (-1)^n/(5n+2) = (1/5)*(sqrt(5)*log(phi) - log(2) + Pi*(5*phi^2)^(-1/4)), where 2*phi=1+sqrt(5).
Sum_{n>=0} (-1)^n/(5n+2) = (1/5)*(sqrt(5)*log(2*sin(3*Pi/10)) - log(2) + (Pi/2)*sec(Pi/10)).
(End)
Sum_{n>=0} (-1)^n/(5n+2) = (Psi(1/5) - Psi(7/10))/10 , see A200135 and A354643. - Robert Israel, Oct 08 2015
From Peter Bala, Feb 19 2024: (Start)
Equals (1/2)*Sum_{n >= 0} n!*(5/2)^n/(Product_{k = 0..n} 5*k + 2) = (1/2)*Sum_{n >= 0} n!*(5/2)^n/A047055(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(5*k + 2)).
Continued fraction: 1/(2 + 2^2/(5 + 7^2/(5 + 12^2/(5 + ... + (5*n + 2)^2/(5 + ... ))))).
The slowly converging series representation Sum_{n >= 0} (-1)^n/(5*n + 2) for the constant can be accelerated to give the following faster converging series
1/4 + (5/2)*Sum_{n >= 0} (-1)^n/((5*n + 2)*(5*n + 7)) and
19/56 + (5^2/2)*Sum_{n >= 0} (-1)^n/((5*n + 2)*(5*n + 7)*(5*n + 12)).
These two series are the cases r = 1 and r = 2 of the general result:
for r >= 0, the constant equals C(r) + ((5/2)^r)*r!*Sum_{n >= 0} (-1)^n/((5*n + 2)*(5*n + 7)*...*(5*n + 5*r + 2)), where C(r) is the rational number (1/2)*Sum_{k = 0..r-1} (5/2)^k*k!/(2*7*12*...*(5*k + 2)). The general result can be proved by the WZ method as described in Wilf. (End)
From Peter Bala, Mar 03 2024: (Start)
Equals (1/2)*hypergeom([2/5, 1], [7/5], -1).
Gauss's continued fraction: 1/(2 + 2^2/(7 + 5^2/(12 + 7^2/(17 + 10^2/(22 + 12^2/(27 + 15^2/(32 + 17^2/(37 + 20^2/(42 + 22^2/(47 + ... )))))))))). (End)

A262178 Decimal expansion of Sum_{k>=0} (-1)^k/(3*k+1)^2.

Original entry on oeis.org

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

Offset: 0

Bruno Berselli, Sep 14 2015

Also, decimal expansion of Sum_{h>=0} Sum_{j=0..h} (-1)^j*binomial(h, j)/(4*(1 + h)*(1 + 6*j)*(2 + 3*j)).

			1 - 1/16 + 1/49 - 1/100 + 1/169 - 1/256 + 1/361 - 1/484 + ...
0.9515177134164150418664828314727415315447285082326970513300324315296113...

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A006752.
Cf. A113476: Sum_{k>=0} (-1)^k/(3*k+1).
Cf. A226735: Sum_{k>=0} (-1)^k/(3*k+1)^3.

RealDigits[(Zeta[2, 1/6] - Zeta[2, 2/3])/36, 10, 100][[1]]

sumalt(k=0, (-1)^k/(3*k+1)^2) \\ Michel Marcus, Sep 14 2015

zetahurwitz(2,1/6)/36 - zetahurwitz(2,2/3)/36 \\ Charles R Greathouse IV, Jan 31 2018

Equals (zeta(2, 1/6) - zeta(2, 2/3))/36, where zeta(s,a) is the Hurwitz zeta function.

cp's OEIS Frontend

A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4*k+1).

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A181049 Decimal expansion of (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+3).

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A193534 Decimal expansion of (1/3) * (Pi/sqrt(3) - log(2)).

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A193535 Decimal expansion of log(2)/3.

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A024217 a(n) = ( Product {k = 1..n} 3*k - 2 ) * ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 2) ).

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A181122 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+1).

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A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).

A024217 a(n) = ( Product {k = 1..n} 3k - 2 ) ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 2) ).