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Ricky
2006-02-28 04:07:59

Oh, and one thing I forgot to mention:

For example we could have the function f(x)=1/x
lim(1/x) when x goes to eternity is 0, RIGHT ?
Qn = f(1) + f(2) + f(3) + f(4) + f(5) + ... + f(n), when n goes to eternity

Now if what you say is correct then Qn goes to 0 ????????

You picked a great example!  As it turns out, even though that limit goes to 0, the sum diverges, that is, goes to infinity.

It is a great example to show that the converse of the 2nd thm I posted:

Is utterly false.

Ricky
2006-02-27 14:48:38

Sorry to all other guys - you were correct with your conclusions but I didnt areed with the way how you got there.

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.

affirmation
2006-02-27 06:57:48

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

Cheers, guys.

ryos
2006-02-27 03:36:26

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.

Look, I'll show you. You can do polynomial division on that function to simplify it, and it comes out as:
x² + 5x + 6 + 1/(x² + x)

...Where you see that the fractional bit goes to 0 as n-> inifinity, but there is still a quadratic, with nothing taking away from it, adding to the sum.

affirmation
2006-02-27 01:29:12

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.

For example we could have the function f(x)=1/x
lim(1/x) when x goes to eternity is 0, RIGHT ?
Qn = f(1) + f(2) + f(3) + f(4) + f(5) + ... + f(n), when n goes to eternity

Now if what you say is correct then Qn goes to 0 ???????? ehehe, I really dont think so. Do some calculus when n=5 and see what you get and after that imagine what happens when n goes to eternity.

Ricky
2006-02-27 01:15:15

affirmation, are you talking about an arbitrary, but non infinite, n?  In your opening post, you wrote:

where we could find the sum of all members.

Which I thought you meant to be as n goes to infinity.

Ansette
2006-02-26 21:42:25

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.

affirmation
2006-02-26 21:26:19

#### Ricky wrote:

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.

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.

Ricky
2006-02-26 07:42:02

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.

affirmation
2006-02-26 06:35:26

#### John E. Franklin wrote:

I don't know, but in general it would be larger than n^2.

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.

John E. Franklin
2006-02-26 04:50:40

I don't know, but in general it would be larger than n^2.

affirmation
2006-02-26 03:50:46

f(x) = (x^4 + 6x^3 + 11x^2 + 6x + 1)/(x*(x+1)) - This is the right way of writing it.
(x^4 + 6x^3 + 11x^2 + 6x + 1) is devided by x*(x+1)

affirmation
2006-02-26 03:45:03

Hi Guys. This problem has taken my mind so much that I cant sleep before solving it. Please help. And here it is -
f(x) = (x^4 + 6x^3 + 11x^2 + 6x + 1)/x*(x+1)
Sn = f(1) + f(2) + f(3) + ... + f(n)
Sn = ?

Thanks