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You are not logged in. #1 20120422 07:59:50
A tough sum  limit?The above problem came up in the now famous gAr series thread. Generally I stay away from that thread. It is for the members to work on and enjoy without me butting in. I was asked for some help on this one and because this one has many interesting points I agreed. Of course before I attempt to answer I must do some ranting. begin Rant(): I believe that this sort of problem is easily handled through the methods I have been going on and on about in this thread for more than a century now. Apparently I can foam at the mouth for as long as I like about how no one is reading any of it. Why is that? Why is there not 1 million replies to each of these problems? Why do posters keep heading over to other forums to read non solutions full of homeomorphisms, iosomorphisms and more rings than a jeweler? Beats me! End Rant: Return(answer) Here is how we can do it using the methods outlined here. These methods allow one to get exact answers to dificult problems using commonsense reasoning. First thing we observe that as n gets larger so does k and so does k*n. That means those 3 constants a,b,c are going to be drowned out. We can reduce the problem to and then to this. This is easily handled by a CAS: Code:n = 300000000000000000000000; NSum[1/Sqrt[k n + n^2], {k, 1, n}, WorkingPrecision > 25] the output is 0.8284271247461900976033770... Code:NSum[Sqrt[k n + n^2 + 31]/( Sqrt[k n + n^2 + 2] Sqrt[k n + n^2 + 5]), {k, 1, n}, WorkingPrecision > 25] the output is 0.8284271247461900976033770... what now? A PSLQ of course! We use one on the above constant and come up with: We have a conjecture, a good one. We are done. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #2 20120422 21:41:02
Re: A tough sum  limit?So,you hqve no definite proof of your answer.Ahhhhh... Well this will do I guess. But I would still like to do it by a hand method. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #3 20120422 21:43:32
Re: A tough sum  limit?Does a hand method automatically mean a proof? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #4 20120422 21:48:57
Re: A tough sum  limit?If it is correct,yes. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #5 20120422 21:49:52
Re: A tough sum  limit?If it is long and difficult how do we verify its correctness? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #6 20120422 21:51:40
Re: A tough sum  limit?How do we verify the correctness of your answer? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #7 20120422 21:55:52
Re: A tough sum  limit?We are sure of the amount of digits 1 that are provided. It is now possible to back engineeer that result. One method suggests itself right away. The point is, it is much better to have 25 digit conjecture than a 0 digit nothing. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #8 20120422 21:59:21
Re: A tough sum  limit?Ok. I think I am gonna try it "my way",but even then I will use your result from computer math to help me. So it is not a waste in the end. Last edited by anonimnystefy (20120422 21:59:40) The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #9 20120422 22:12:57
Re: A tough sum  limit?Do you have a suggestion for a hand method? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #10 20120422 22:21:20
Re: A tough sum  limit?I am sorry, I do not. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #11 20120423 07:01:07
Re: A tough sum  limit?Okay. Thank,anyways,for this method here. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #12 20120423 07:03:44
Re: A tough sum  limit?I think my earlier comments about a gAr problem apply here doubly. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 