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Suppose we have the series $$\sum\limits_{n=1}^{\infty} \frac{n+2}{(n+1)\sqrt{n+3}}$$ and want to test for convergence. I have tried a number of things--ratio test, various comparison tests, Divergence Test, etc.--and can't seem to show that this series diverges. If someone could point me in the right direction on this, I would greatly appreciate it.

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  • $\begingroup$ Short imprecise answer: Look only at the highest ordered terms of the numerator and denominator. $\sum\limits_{n=1}^\infty \frac{n+2}{(n+1)\sqrt{n+3}}$ acts similarly to $\sum\limits_{n=1}^\infty \frac{n}{n\sqrt{n}}$ which you should know more about. $\endgroup$ – JMoravitz May 25 '18 at 17:54
  • $\begingroup$ @Matt.P Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Jun 22 '18 at 20:36
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Simply note that

$$\frac{n+2}{(n+1)\sqrt{n+3}}\sim \frac1{\sqrt n}$$

then refer to limit comparison test.

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hint: Compare the series with $\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}$

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Have you tried limit comparison with

$$\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}} ?$$

The series is divergent in comparison with a $p$ series with $ p<1$

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In other approach, I looked for the value of the sum accordance with the following:

$\sum\limits_{n=1}^{\infty} \frac{n+2}{(n+1)\sqrt{n+3}}=\sum\limits_{n=2}^{\infty} \frac{n+1}{n\sqrt{n+2}}=\sum\limits_{n=2}^{\infty} \frac{n}{n\sqrt{n+2}}+\sum\limits_{n=2}^{\infty} \frac{1}{n\sqrt{n+2}}=$

$\sum\limits_{n=4}^{\infty} \frac{1}{\sqrt{n}}+\sum\limits_{n=2}^{\infty} \frac{1}{n\sqrt{n+2}}$

where $\sum\limits_{n=4}^{\infty} \frac{1}{{n^{\frac{1}{2}}}}=\zeta({\frac{1}{2}})-(1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3})\rightarrow \infty$

and $\sum\limits_{n=2}^{\infty} \frac{1}{n\sqrt{n+2}}\lt \sum\limits_{n=2}^{\infty} \frac{1}{n\sqrt{n}}\rightarrow \zeta({\frac{3}{2}})-\frac{1}{2}\approx 2.1124$ $\big(\zeta(s)$ is the Riemann zeta function $\big)$

So the series is divergent.

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