Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #26 20120723 03:45:44
Re: a challenging problem for allIf you meant an expression for the closed form then it is obviously not possible. The sum will either equal a constant or infinity. 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. #27 20120723 03:47:03
Re: a challenging problem for allIt is possible. 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 #28 20120723 03:49:48
Re: a challenging problem for allWith a variable as the upper index yes. But with an infinity up there the sum is either a constant like 1 / 2 ( just picked as an example ) or ∞. 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. #29 20120723 03:52:02
Re: a challenging problem for allIt can be an expression in terms of x. 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 #30 20120723 07:01:52
Re: a challenging problem for allYes, it could. 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. #31 20120815 23:05:45
Re: a challenging problem for allhi bobbym any information about my question #32 20120815 23:35:48
Re: a challenging problem for allHi rajinikanth0602; By the ratio test: By the ratio test the series diverges for all values of x. 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. #33 20120902 19:17:30
Re: a challenging problem for alli think post reply ''19'' is not correct #34 20120902 19:18:23
Re: a challenging problem for allHi;
In exchange for the rupees I would like your acknowledgement that the problem has been solved with the proof that is in post#32, or do you require more? 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. #35 20120902 21:05:52
Re: a challenging problem for allHi rajinikanth0602 Last edited by anonimnystefy (20120902 21:06:11) 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 