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You are not logged in. #1 20060226 03:45:03
DesperateHi Guys. This problem has taken my mind so much that I cant sleep before solving it. Please help. And here it is  #2 20060226 03:50:46
Re: Desperatef(x) = (x^4 + 6x^3 + 11x^2 + 6x + 1)/(x*(x+1))  This is the right way of writing it. #3 20060226 04:50:40
Re: DesperateI don't know, but in general it would be larger than n^2. igloo myrtilles fourmis #4 20060226 06:35:26
Re: Desperate
I think you are wrong. If what you are saying is correct  Sn>n^2 => Sn equals eternety because n goes to eternety. I think that f(1), f(1) + f(2), f(1) + f(2) +f (3) etc or f(1), f(2), f(3)/sperately/ should be proved to be part of arithmetical or geometrical progression where we could find the sum of all members. If they are not such progressions I have no idea how to solve this problem. #5 20060226 07:42:02
Re: Desperate
Last edited by Ricky (20060226 07:44:58) "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #6 20060226 21:26:19
Re: Desperate
Cheers Ricky but I think you are not right. The limes of f(x) when n goes to eternity is no reason of concluding that that the sum of f(1) + f(2) + f(3) + ... + f(n) is eternity. You are not considering that with each new f(n) added to the sum the dividends is growing unimaginable large. #7 20060226 21:42:25
Re: Desperatewell the point being is that it's the sum to infinity, with the terms ever approaching x² (with use of the limit, which is a standard point) then each f(x) is approaching x² and thus it's a divergant series adding up to a sum that approaches infinity as x>infinity. #8 20060227 01:15:15
Re: Desperateaffirmation, are you talking about an arbitrary, but non infinite, n? In your opening post, you wrote:
Which I thought you meant to be as n goes to infinity. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #9 20060227 01:29:12
Re: DesperateRicky  yes n is going to eternity but the sum does not go so far. Ansette please post mathematical conclusions not what you think, because you think wrong. #10 20060227 03:36:26
Re: DesperateLook at it this way, affirmation. The degree of the denominator is less than that of the numerator. As n> infinity, the top gets much larger than the bottom, such that the division becomes insignificant, and the sum goes to infinity. El que pega primero pega dos veces. #11 20060227 06:57:48
Re: DesperateThank you for your reply ryos. I've never thought of it that way. Sorry to all other guys  you were correct with your conclusions but I didnt areed with the way how you got there. ryos thanks again, this was one of the problems I was given on a Math competition  the only one I couldn't solve. Shame on me #12 20060227 14:48:38
Re: Desperate
Whether you agree or not, both of our methods are correct. There are theorems that say: When n > m. Actually, that one I believe I can prove. There is also a theorem that says if: Then the sum of the series is divergent. Since infinity does not equal 0, both mine and Ansette's conclusions are not only correct, but well reasoned. Last edited by Ricky (20060227 14:49:52) "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #13 20060228 04:07:59
Re: DesperateOh, and one thing I forgot to mention:
You picked a great example! As it turns out, even though that limit goes to 0, the sum diverges, that is, goes to infinity. Is utterly false. Last edited by Ricky (20060228 04:08:44) "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." 