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 February 26th, 2018, 12:53 PM #1 Newbie   Joined: Aug 2017 From: Norway Posts: 10 Thanks: 0 Series Hey! I'm stuck on this task about series. Determine whether the following series converges or diverges: $\displaystyle \sum\limits_{n=0}^\infty \frac{n}{n^2 \sqrt n +1}$ I've tried to use the ratio test. All help much appreciated. Last edited by deltaX; February 26th, 2018 at 01:00 PM.
 February 26th, 2018, 01:39 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,767 Thanks: 1422 note $\displaystyle \sum_{n=0}^\infty \dfrac{n}{n^2\sqrt{n}+1} = \sum_{n=1}^\infty \dfrac{n}{n^2\sqrt{n}+1}$ further ... $\dfrac{n}{n^2\sqrt{n}+1} = \dfrac{1}{n\sqrt{n}+\frac{1}{n}} < \dfrac{1}{n^{3/2}}$ note $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{3/2}}$ is a convergent p-series Thanks from deltaX
February 26th, 2018, 02:08 PM   #3
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 Originally Posted by skeeter note $\displaystyle \sum_{n=0}^\infty \dfrac{n}{n^2\sqrt{n}+1} = \sum_{n=1}^\infty \dfrac{n}{n^2\sqrt{n}+1}$ further ... $\dfrac{n}{n^2\sqrt{n}+1} = \dfrac{1}{n\sqrt{n}+\frac{1}{n}} < \dfrac{1}{n^{3/2}}$ note $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^{3/2}}$ is a convergent p-series
Which explains why the ratio test didn't work. The ratio test is just a fancy way of doing a comparison with a geometric series. It can't do comparisons with series as fine as the p-series.

Note that if your series is a fraction of terms in n to some (possibly fractional) power, then a comparison will always work. The limit comparison test is there probably the easiest.

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