You are not logged in.

- Topics: Active | Unanswered

**affirmation****Guest**

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

**affirmation****Guest**

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)

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

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

**igloo** **myrtilles** **fourmis**

Offline

**affirmation****Guest**

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.

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

*Last edited by Ricky (2006-02-25 08:44:58)*

"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..."

Offline

**affirmation****Guest**

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.

**Ansette****Member**- Registered: 2006-02-19
- Posts: 21

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.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"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..."

Offline

**affirmation****Guest**

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.

Cheers, guys, but your conclusions arent correct. Please someone helps here.

Thanks for your replies all.

**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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.

El que pega primero pega dos veces.

Offline

**affirmation****Guest**

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.

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

*Last edited by Ricky (2006-02-26 15:49:52)*

"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..."

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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 eternityNow 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.

*Last edited by Ricky (2006-02-27 05:08:44)*

Offline