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October 22nd, 2018, 09:11 AM   #1
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Convergence Series

Show whether the series converges or diverges
$\displaystyle \sum \limits_{n=1}^{\infty} \left(\frac{1+3+3^2+...+3^n}{3^n + 3^{n-1} }\right)^n$

Last edited by skipjack; October 22nd, 2018 at 01:38 PM.
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October 22nd, 2018, 10:12 AM   #2
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$\displaystyle \lim \limits_{n\to\infty}~\frac{\sum \limits_{k=0}^n~3^k}{3^n+3^{n-1}} = \frac 9 8 > 1$

So the root test says this series diverges.

Last edited by skipjack; October 23rd, 2018 at 05:24 PM.
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October 22nd, 2018, 02:29 PM   #3
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The fraction is always > 1, so there is no way for the series to converge.
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October 23rd, 2018, 12:56 AM   #4
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Quote:
Originally Posted by mathman View Post
The fraction is always > 1, so there is no way for the series to converge.
In fact, the fraction itself converges to 27/24 which is absolutely >1.

Appendum: I suppose that's the outcome of ROMSEKs post as well, but I couldn't read it since I got a dfrac-error message.

Last edited by Arisktotle; October 23rd, 2018 at 01:37 AM.
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October 23rd, 2018, 01:26 AM   #5
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I've changed it to use frac instead of dfrac.
Thanks from greg1313
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October 23rd, 2018, 01:40 AM   #6
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@skipjack. Thanks, but 1 dfrac instance left in romsek's post.

I get these messages regularly. Anything I can do to overcome them?

Last edited by skipjack; October 23rd, 2018 at 05:30 PM.
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October 23rd, 2018, 02:20 PM   #7
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Quote:
Originally Posted by Arisktotle View Post
In fact, the fraction itself converges to 27/24 which is absolutely >1.

Appendum: I suppose that's the outcome of romsek's post as well, but I couldn't read it since I got a dfrac-error message.
The question was about the series, not the individual terms.

Last edited by skipjack; October 23rd, 2018 at 05:31 PM.
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