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February 26th, 2018, 12:53 PM   #1
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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.
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February 26th, 2018, 01:39 PM   #2
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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
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February 26th, 2018, 02:08 PM   #3
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Originally Posted by skeeter View Post
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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