I want to show that $\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $ converges or diverges. I want to show that $$\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $$ converges (absolutly?) or diverges.
My idea was: $n=2k+1$ and then it becomes: $$\sum_{n=1}^\infty (-1)^n\cdot\frac{(2k+1)^2+2}{(2k+1)^3+2k+1} $$
Then I somehow try to show whether the limit is $0$ and whether it is decreasing. But I am not sure how I would go about that?

You can show the terms are decreasing by finding $a_{n}-a_{n+1}$ and showing it's positive:
$\dfrac{n^2+2}{n^3+n}-\dfrac{(n+1)^2+2}{(n+1)^3+(n+1)} = \dfrac{n^4+2n^3+6n^2+5n+4}{n(n+1)(n^2+1)(n^2+2n+2)}>0$
Regarding the limit:
$\dfrac{n^2+2}{n^3+n} = \dfrac{1+\frac{2}{n^2}}{n+\frac{1}{n}}$. So as n goes to infinity the numerator goes to 1, and the denominator goes to infinity. So the limit is 0. So the series converges by the alternating series test. 
As far as absolute vs conditional convergence:
$\dfrac{n^2+2}{n^3+n} = \dfrac{n^2}{n^3+n} + \dfrac{2}{n^3+n}$
So we can think of this as the sum of 2 series. The second converges.
Regarding the first:
$\dfrac{n^2}{n^3+n} = \dfrac{1}{n+\frac{1}{n}} >= \dfrac{1}{n+1}$ for n>=1
$\dfrac{1}{n+1}$ diverges by the integral test. Our series is greater so it also diverges.
Sum of a divergent and convergent series is divergent. So the absolute series diverges. So our original series converges conditionally.
