4

I am trying to find

$$\frac{4}{9} + \frac{5}{27} + \frac{7}{81} + \frac{8}{243} + \frac{13}{729} + \frac{15}{2187} + \frac{31}{6561} + \frac{33}{19683} + \frac{34}{59049} + \cdots $$

I have tried to let the sum be $S$ then multiply by $3$ and subtract it from itself but I am not getting it to work. Please give a hint or solution thank you :)

Αδριανός asks about numerators so I will describe how I was finding them. For each number $a$ I check if there is prime $b$ so that $$(a-1)!-1=b$$

If there is then $a$ becomes next numerator. If not then skip, etc.

For example if $a=13$ then there is prime $b$:

$$(13-1)!-1=479001599 = b$$

terrace
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    Well, I'm trying to figure out what is going on with the numerators...the answer may lay there – KR136 Apr 10 '16 at 00:57
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    The sequence isn't in OEIS, but maybe it is a play on a well-known one. – KR136 Apr 10 '16 at 01:01
  • The only discernible pattern amongst the numerators I can see, is that, aside from the difference between $\frac{1}{3}$ and $\frac{4}{9}$, the difference between the numerators of the terms follows the repetitive pattern $1, 2, 1, 5, 1, 2, 1, 5...$ – KR136 Apr 10 '16 at 01:03
  • Since it seems to have period 4, I think multiplying by $81(=3^4)$ and subtracting is a better idea. – Element118 Apr 10 '16 at 01:04
  • @Αδριανός Sorry I was wrong after the first 6 numbers and I am still not sure about the first term. The numerator are number $a$ so that there is a prime $b$ with $$(a-1)!-1 = b$$ I am checking to find them only by brute force in mathematica so I don't know if there is a easy way to write it. – terrace Apr 10 '16 at 01:11
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    If you cannot give the full list of numerators, how do you expect to solve the sum? – YoTengoUnLCD Apr 10 '16 at 01:12
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    @YoTengoUnLCD You cannot give full list of digits of pie, but you can write equations with pie. That mean you can maybe find sum without actually knowing every number only how to find every number. – terrace Apr 10 '16 at 01:13
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    Mmm...digits of pie. ¶ @terrace: You should put the property of the numerators (numbers $a$ such that $(a-1)!$ is one more than a prime) in the problem, rather than burying it in the comments. Why are you interested in this series? Without any further context, it does seem a little contrived. – Brian Tung Apr 10 '16 at 01:16
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    By the way, @terrace have you looked into Wilson's theorem...I'm trying to figure out if there might be a way to relate your relationship and that definition of primality – KR136 Apr 10 '16 at 01:16
  • Ok, can you calculate $\sum_{n>0} a_n$ where $a_1=1, a_i=? (i=2,3,...)$? – YoTengoUnLCD Apr 10 '16 at 01:17
  • @Αδριανός BrianTung I put it in the quetion – terrace Apr 10 '16 at 01:17
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    @Αδριανός I think it would be better for if it was $(a-1)!+1=b$ but I dont know how to do it for $(a-1)!-1=b$ – terrace Apr 10 '16 at 01:18
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    See here. The fact that there are no more known indicates that your problem is open at the moment. – Brian Tung Apr 10 '16 at 01:20
  • Note that $a=$round$(\frac{L(b+1)}{W(\frac{L(b+1)}{e})}+\frac{1}{2})$ where $W(x)$ is the Lambert-W function, and the function $L(x)$ is described in greater detail at this page: http://mathforum.org/kb/message.jspa?messageID=342551&tstart=0 – KR136 Apr 10 '16 at 01:30
  • @Αδριανός What is L and W? Is e =2.71828128...? edit: ok – terrace Apr 10 '16 at 01:31
  • @terrace Sorry I posted prematurely. Yes $e=2.718...$. See the link for detail on the construction of the function, but basically it is an approximation (hence the rounding) of the inverse of the gamma function. Though I know links are not encouraged on this site, the approximation in question is already featured on MathOverflow, and I did not want to take the pains to transcribe it again here, because it is not the object of the question. – KR136 Apr 10 '16 at 01:32
  • @terrace The issue remains however, that $b$ must be a prime and $a$ be an integer, so it doesn't really take us anywhere, and in fact might be misleading because of the rounding. – KR136 Apr 10 '16 at 01:34
  • Note that even if you get a general term, summing the series may not be easier. Unless you are interested in a closed form, summing the first 100 terms may get you a good estimate since the denominator is growing wild. – NoChance Apr 10 '16 at 01:35
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    the numerators are A002982(n)+1. – AccidentalFourierTransform Jun 06 '17 at 21:59
  • @AccidentalFourierTransform Wow, thank you – terrace Jun 10 '17 at 23:12

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