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Oh, 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.
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.
Thank 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
Look 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.
Ricky - 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.
affirmation, 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.
well 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.
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.
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.
I don't know, but in general it would be larger than n^2.
f(x) = (x^4 + 6x^3 + 11x^2 + 6x + 1)/(x*(x+1)) - This is the right way of writing it.
Hi Guys. This problem has taken my mind so much that I cant sleep before solving it. Please help. And here it is -