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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2006-01-12 09:03:13

I think i see, thanks. But i'm dealing with the sequence, not the sum of the infinite series. I wasn't too clear about the difference before, but i am now. Sorry for the misunderstaning smile

2006-01-12 08:59:39

Think of it this way:

Where c is not 0, then basically what you are doing is:

Any number that you add an infinite amount of times goes to infinity.  And since An approaches this c, you are adding it an infinite amount of times.

Edit: God, I'm starting to think you're right.  I've gotten into a habit of hearing converge and diverge, and automatically thinking infinite series.  The An is also common when doing infinite series.

Wonder why (s)he didn't object to my restating of the question using summation though...

Edit #2:

It sorta sounds silly, but only because you misunderstood the question.

Starting to find that quite ironic right now.

2006-01-12 08:58:04

(n^4 + 2n^2) / (4n^4 + 7n)

Rick, I think he's dealing with the sequence a_n = (n^4 + 2n^2) / (4n^4 + 7n), not the sum of the infinite series.

So if that is the case, your method is correct, by dividing the top and bottom by n^4, you find that as n -> infinity, the value of a_n approaches 1/4.

If you are in fact dealing with the sum of the infinite series, keep in mind that as you are approaching a_n -> 1/4, you're adding a whole bunch of (more specifically, an infinite ammount of) 1/4's.

So the sequence converges but the series does not.

2006-01-12 08:20:34

I'm a bit confused by the converge/diverge thing. I thought that a sequence diverges when r>1, so it doesn't tend to any one limit?

2006-01-12 07:38:49

Not quite.  An heads to 1/4 as n heads to infinity, that part was right.  But this means the series diverges.

2006-01-12 07:12:52

Ah yes, thanks, i've figured it.

Divide through by n^4 which leaves (1 + 1/n^2) / (4 + 7/n^3) . Since 1/n^2 and 7/n^3 both converge to zero, this means the series must converge to the limit 1/4.  Back of the net.

2006-01-11 14:16:43

It sorta sounds silly, but only because you misunderstood the question.

His question is:

First, you have to find whether it converges or diverges.  If you have a sequence An, and:

Then the series diverges.  Try to take this limit.

2006-01-11 10:32:20

This might sound silly, but did you try to take the derivative and set it to zero?

2006-01-11 09:25:50

Does anyone happen to know how i could go about finding what this sequence converges to?

An = (n^4 + 2n^2) / (4n^4 + 7n)

I can manage simpler ones but this one's causing me a few headache... just a hint would be nice :-)


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