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.

]]>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.

]]>Cheers, guys.

]]>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.

]]>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.

]]>where we could find the sum of all members.

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

]]>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.

]]>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.

]]>

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.

]]>(x^4 + 6x^3 + 11x^2 + 6x + 1) is devided by x*(x+1)]]>

f(x) = (x^4 + 6x^3 + 11x^2 + 6x + 1)/x*(x+1)

Sn = f(1) + f(2) + f(3) + ... + f(n)

Sn = ?

Thanks

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