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 October 26th, 2013, 11:40 AM #1 Newbie   Joined: Mar 2013 Posts: 24 Thanks: 0 Nth term test for series The series is the sum of (3 * ?n) / ln(9n) where n = 2 and goes to infinite. So far, we've only be taught to do the integral test, Nth term test, and to check for geometric or telescopic series. The only on that applies here is the Nth term test, but after working it out, I found the lim(n -> inf) of the Nth term to be 0, meaning it is indeterminate from this test. Can anyone help me understand why this is series is divergent?
October 26th, 2013, 01:00 PM   #2
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Re: Nth term test for series

Quote:
 Originally Posted by Polaris84 The series is the sum of (3 * ?n) / ln(9n) where n = 2 and goes to infinite. So far, we've only be taught to do the integral test, Nth term test, and to check for geometric or telescopic series. The only on that applies here is the Nth term test, but after working it out, I found the lim(n -> inf) of the Nth term to be 0, meaning it is indeterminate from this test. Can anyone help me understand why this is series is divergent?
The sequence $\frac{3 \sqrt{n}}{ln(9n)}$ diverges!

 October 26th, 2013, 01:38 PM #3 Newbie   Joined: Mar 2013 Posts: 24 Thanks: 0 Re: Nth term test for series Okay, so is this the correct usage of L'Hopital's rule? $lim(n\rightarrow\infty)\frac{3 \sqrt{n}}{ln(9n)}= lim(n\rightarrow\infty)\frac{\frac{3}{2\sqrt{n}}}{ \frac{1}{n}} = lim(n\rightarrow\infty)\frac{3n}{2\sqrt{n}} = lim(n\rightarrow\infty)\frac{3}{2}\sqrt{n} = \infty$ Because the first time I did the problem I used L'Hopital's again after simplifying to: $lim(n\rightarrow\infty)\frac{3n}{2\sqrt{n}}= lim(n\rightarrow\infty)-\frac{3}{n^{\frac{3}{2}}} = 0$ or something like that.
October 26th, 2013, 02:12 PM   #4
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Re: Nth term test for series

Quote:
 Originally Posted by Polaris84 Okay, so is this the correct usage of L'Hopital's rule? $lim(n\rightarrow\infty)\frac{3 \sqrt{n}}{ln(9n)}= lim(n\rightarrow\infty)\frac{\frac{3}{2\sqrt{n}}}{ \frac{1}{n}} = lim(n\rightarrow\infty)\frac{3n}{2\sqrt{n}} = lim(n\rightarrow\infty)\frac{3}{2}\sqrt{n} = \infty$
Yes

Quote:
 Originally Posted by Polaris84 Because the first time I did the problem I used L'Hopital's again after simplifying to: $lim(n\rightarrow\infty)\frac{3n}{2\sqrt{n}}= lim(n\rightarrow\infty)-\frac{3}{n^{\frac{3}{2}}} = 0$
Should have been:

$lim(n\rightarrow\infty)\frac{3n}{2\sqrt{n}}= lim(n\rightarrow\infty)\frac{3}{n^{-\frac{1}{2}}} = 3\sqrt{n}$

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