A257964 Decimal expansion of Sum_{n=1..infinity} (-1)^(n-1)/(n + log(n)).

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

Offset: 0

Author

Iaroslav V. Blagouchine, May 14 2015

Keywords

• nonn
• cons

Comments

This alternating series converges relatively slowly, but can be efficiently computed via an integral representation, which converges exponentially fast (see my formula below). I used this formula and PARI to compute 1000 digits of this series. Modern CAS are also able to evaluate it very quickly and to a high degree of accuracy.

Examples

			0.769402150280800484122126971946005315576205532033543...
		

Links

• Iaroslav V. Blagouchine, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A099769, A257837, A257812, A257898, A257960, A257972.

Programs

Maple

evalf(sum((-1)^(n-1)/(n+ln(n)), n = 1..infinity), 120);
evalf(1/2+int((x+arctan(x))/(sinh(Pi*x)*((1+(1/2)*ln(1+x^2))^2+(x+arctan(x))^2)), x = 0..infinity), 120);

Mathematica

N[NSum[(-1)^(n-1)/(n+Log[n]), {n, 1, Infinity}, AccuracyGoal -> 120, Method -> "AlternatingSigns", WorkingPrecision -> 200],120]
N[1/2 + NIntegrate[(x+ArcTan[x])/(Sinh[Pi*x]*((1+1/2*Log[1+x^2])^2 + (x+ArcTan[x])^2)), {x, 0, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200],120]

PARI

default(realprecision, 120); sumalt(n=1, (-1)^(n-1)/(n+log(n)))

PARI

allocatemem(50000000);
default(realprecision, 1200); 1/2 + intnum(x=0, 1000, (x+atan(x))/(sinh(Pi*x)*((1+0.5*log(1+x^2))^2 + (x+atan(x))^2)))

Formula

Equals 1/2 + integral_{x=0..infinity} (x+arctan(x))/(sinh(Pi*x)*((1+1/2*log(1+x^2))^2 + (x+arctan(x))^2)).