Question on Series Convergence 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. 
 A: Simply note that
$$\frac{n+2}{(n+1)\sqrt{n+3}}\sim \frac1{\sqrt n}$$
then refer to limit comparison test.
A: hint: Compare the series with $\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}$
A: 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$
A: 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.
