A190303 Decimal expansion of sum of alternating series of reciprocals of Ramanujan primes, Sum_{n>=1} (1/R_n)(-1)^(n-1), where R_n is the n-th Ramanujan prime, A104272(n).

4, 4, 6, 6, 8, 4, 3, 0, 7

Offset: 0

Author

John W. Nicholson, May 07 2011

Keywords

• nonn
• cons
• more

Comments

Computed 0.446684 for n = 1 to 65536, using Open Office Calc. Next digit expected to be between 2 and 3.

By computing all Ramanujan primes less than 10^9, we find that about 9 decimal places of the sum should be correct: 0.446684307 (truncated, not rounded). The following table shows the number of Ramanujan primes between powers of 10 and the sum of the alternating reciprocals of those primes.

1 1 0.50000000000000000

2 9 -0.05765566386047510

3 62 0.00388002010130731

4 487 0.00050881775862179

5 3900 -0.00004384563815649

6 32501 -0.00000552572415587

7 279106 0.00000045427780897

8 2444255 0.00000005495474474

9 21731345 -0.00000000549864067

Total: 0.44668430669928564 - T. D. Noe, May 08 2011

Let E_n denote the error after the first n terms in the series. Then by the Alternating Series Test, 1/R_{n+1} - 1/R_{n+2} < E_n < 1/R_{n+1}. [Jonathan Sondow, May 10 2011]

Examples

			0.446684307...
		

Links

• J. Sondow, Ramanujan primes and Bertrand's postulate, arXiv:0907.5232 [math.NT], 2009-2010.
• J. Sondow, Ramanujan primes and Bertrand's postulate, Amer. Math. Monthly 116 (2009), 630-635.
• J. Sondow, J. W. Nicholson, and T. D. Noe, Ramanujan Primes: Bounds, Runs, Twins, and Gaps, J. Integer Seq. 14 (2011) Article 11.6.2
• Eric W. Weisstein, Prime Sums

Crossrefs

Cf. A104272, A085548, A078437, A190124.

Formula

Sum_{n>=1} (-1)^(n-1)(1/R_n), where R_n is the n-th Ramanujan prime, A104272(n).

Extensions

Definition corrected by Jonathan Sondow, May 10 2011