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A190303
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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).
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1
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OFFSET
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0,1
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COMMENTS
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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]
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LINKS
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FORMULA
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Sum_{n>=1} (-1)^(n-1)(1/R_n), where R_n is the n-th Ramanujan prime, A104272(n).
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EXAMPLE
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0.446684307...
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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