I would like to know whether this converges (like the sum of the reciprocals of all twin primes) or diverges (like the sum of the reciprocals of all primes). If it converges, what does it converge to? And how do we know that those series converge and diverge?
2006-12-01 14:24:04 · 6 answers · asked by Robert S 1 in Science & Mathematics ➔ Mathematics
First of all observe that the:
1st prime number is 1,
2nd prime number is 2,
3th prime number is 3,
4th prime number is 5 and etc.
As the value of the prime number approaches infinity, it's reciprocals approaches zero. So the question is at what point do you stop adding the reciprocals? The number of reciprocals add together will dictate your converge value.
= 1/2 + 1/3 + 1/5 + 1/7 +1/11 + 1/13 + 1/17 + 1/19 + ...... + 1/∞
= 0∙5 + 0∙33..+ 0∙2 + 0∙142..+ 0∙090..+ 0∙076..+ 0∙058..+ 0∙052.+.0
= 1∙4554............
2006-12-01 15:37:13 · answer #1 · answered by Brenmore 5 · 0⤊ 1⤋
The sum of the reciprocals of the twin primes converges? I did not know that.
2006-12-01 14:35:52 · answer #2 · answered by modulo_function 7 · 0⤊ 0⤋
That problem was solved by Euler in the 18th century. The sum of reciprocals of primes diverges, actually the sum of reciprocals of primes less than n is equivalent to ln(ln(n)).
2006-12-02 00:36:18 · answer #3 · answered by frank m 2 · 0⤊ 0⤋
Sorry not an answer, but an observations which may help explain issues to others.
The sum of reciprocals of primes diverges, that means foar any number you care to pick if you pick enough primes then the sum of reciprocals will eventually exceed that number.
2006-12-01 19:27:07 · answer #4 · answered by crazy_tentacle 3 · 0⤊ 0⤋
I see Dragan is first out of the gate with an answer, yet permit's crucially recondition your expression to a greater appropriate type which will enable arbitrary accuracy: a million/? = (a million ? a million/p?) (a million ? a million/p?) (a million ? a million/p?) …(a million ? a million/pn) ? = (p?/(p?-a million)) (p?/(p? -a million)) (p?/(p?-a million)) …(pn/(pn - a million)) Taking logs does the needed portion of changing this product right into a sum... (might besides use organic log): ln(?) = ln((p?-a million)/p?) + ln((p? -a million)/p?) + ln ((p?-a million)/p?) … + ln((pn - a million)/pn) call that: ln(?) = f(p?) + f(p?) + f(p?) … + f(pn) Now we are able to tabulate f(p_k) = ln((p_k -a million)/p_k) and then use Gauss' technique of blunders accumulation to attain arbitrary accuracy. observe that asymptotically f(p_k) would be an extremely small constructive volume, so which you will desire arbitrary-precision. (you may actual get a greater direct Taylor sequence expression than comparing ln((p_k -a million)/p_k) naively as written (in any different case issues destroy down previous f(1e16) ? 0), and that i'm constructive there additionally could be tight asymptotic larger and decrease bounds for ln((p_k -a million)/p_k). I crunched the actual numbers out in Python and my answer so some distance (making use of in basic terms unmarried-precision) is likewise [2, 3, 23, 601,1800589, (any further words are > 1e16)] earlier I prepare the Gauss blunders accumulation technique... TBD [EDIT: quite thrilling posts from the two one among you and Vikram... that's the spirit of collaborative gaining expertise of that Y!A is all approximately]
2016-12-13 18:23:52 · answer #5 · answered by Anonymous · 0⤊ 0⤋
an interesting question! it's amazing you got nonsense for an answer to a truly exciting question!
2006-12-01 16:40:00 · answer #6 · answered by grand_nanny 5 · 0⤊ 1⤋