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Given the probability of flipping a heads with a fair coin is \frac{1}{2}, what is the probability that the first heads occurs on a prime number?
\sum_{n=1}^\infty \left(\frac{1}{2}\right)^{p_n}
where p_n is the nth prime number.
Which gives a value of about 0.41468
Let {p_i} be the sequence of primes. The probability you're looking for would be:
\sum _{i=1} ^{\infty} 0.5^{p_i}
Yes, I realize that the answer is this summation (\sum _{primes} ^{} 1/2^p), which clearly converges very quickly (to about .4146825...). Anyone have any ideas on whether the answer is irrational, or even expressable as a fraction of constants (like \sum _{n=1} ^{\infty} 1/n^2. Anyone know anything else about f(x)=\sum _{primes} ^{} 1/x^p?
I don't really know what more you want, if we wee able to ask your question precisely we'd be to busy polishing our Fields medals to post on Physics Forums.
We can say the series is convergant, but I don't think there's too much more we can say.
There are alot of products and sums involving the nth primes that converge very quickly to a value, there are even somewhoe precise value can be known, but I'm pretty ceratin that this isn't one of them.
Gokul43201
Jun9-04, 05:20 PM
If you had posted this question several centuries ago, I might have said:
"the sum approaches \sqrt{2} - 1 but there is not enough room in this forum for me to prove it. "
and gotten away with that !
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