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» Hatrack River Forum » Active Forums » Books, Films, Food and Culture » Fun with Calculus (Calc 2--series convergence)

   
Author Topic: Fun with Calculus (Calc 2--series convergence)
Starsnuffer
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Ok. The sum from 1 to infinity= (3^(n-1)+1)/3^n
Determine whether it converges or diverges.

I've attempted to do the ratio test: lim n->infinty An+1/An And gotten to (3^n+1)/(3^(n-1)+1)*(1/3). I feel confident that up till then it makes sense/i have it right, but I don't know how to get an answer other than infinty/infinty which is not helpful.

Anyone got an idea for it?

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fugu13
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What other tests have they taught you?
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If you multiply through by the 1/3 you get lim n-> infinity of (3^n+1)/(3^n+3) which is 1 so the test is inconclusive.
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swbarnes2
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quote:
Originally posted by Starsnuffer:
Ok. The sum from 1 to infinity= (3^(n-1)+1)/3^n
Determine whether it converges or diverges.

I've attempted to do the ratio test: lim n->infinty An+1/An And gotten to (3^n+1)/(3^(n-1)+1)*(1/3). I feel confident that up till then it makes sense/i have it right, but I don't know how to get an answer other than infinty/infinty which is not helpful.

Anyone got an idea for it?

I think the sum is infinity. As N gets bigger, the value of the whole thing approaches 1/3. So adding 1/3 over and over again should yield infinity.

Now, if your whole value were approaching 0 as N got big, then you might have a finite sum. But unless you wrote it wrong, or I'm really reading it wrong, that's not happening here. If I had to guess, I'd guess that the sum is never going to be finite as n approaches infinity while once side has something^n, and the other side has something^n-1.

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You can actually just use a comparison test. We know that the sum from 0 to infinity of 1/3 diverges and we know that (3^(n-1)+1)/3^n is greater than 1/3 for all n so the sum of (3^(n-1)+1)/3^n from 1 to infinity diverges.

EDIT: Basically what swbarnes said

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Eisenoxyde
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deleted because I was a moron giving bad info

[ January 24, 2008, 09:34 PM: Message edited by: Eisenoxyde ]

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That's not actually an accurate test. The sum of 1/n from 1 to infinity does not converge and the denominator is always greater than the numerator.
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Eisenoxyde
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Have you discussed L'Hopital's rule in your class yet? This is the first thing I always try when I encounter a series that goes to infinity/infinity.

http://en.wikipedia.org/wiki/L%27Hospital%27s_rule
(Hatrack isn't allowing me to link it for some reason.)

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Eisenoxyde
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Umm, the sum of 1/n from 1 to infinity does converge...
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Then tell me what it converges to.

EDIT: Search for the harmonic series on wikipedia

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Eisenoxyde
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You're right. Sorry, I don't know what I was smoking...
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I think you were thinking of limits instead of sums [Smile]
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Eisenoxyde
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I was thinking about series convergence and not the sums. And now I think it's time for me to take a break from studying before I make any more dumb mistakes.
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Starsnuffer
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Ok thanks guys, moral of the story is : I get that 1 times 1/3=1/3 and adding that gets infinity, so the series is divergent.

but.. the powers aren't quite the same, but is the difference just so small that it doesn't matter. Because otherwise you have ~1/n *1/3 which is 0*1/3 which is 0 and means you have to try something else.

Why isn't 3^(n-1)/3^n=0 as n-> infinty?

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(3^(n-1))/3^n is 1/3 for all n because the powers cancel. 3^(n-1) * 3 = 3^n
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Starsnuffer
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Oh riiight. Wow. That makes everything so much nicer. (grumble about forgetting that rule)
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