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?

The sum converges by the alternating series test. 

It turns out that the sum may be evaluated using complex analysis, i.e., the residue theorem.  First simplify the sum 
$$\sum_{n=0}^{\infty} \sin{\left (n \frac{\pi}{2} \right )} \frac{n^2+2}{n^3+n} $$
as 
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \left [1+\frac1{ (2 k+1)^2 + 1} \right ]  = \frac{\pi}{4} + \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \frac1{ (2 k+1)^2 + 1}$$
Note that the sum on the RHS is equal to
$$\frac12 \sum_{k=-\infty}^{\infty} \frac{(-1)^k}{2 k+1} \frac{1}{ (2 k+1)^2 + 1}  $$
We may evaluate this sum using a result that is a consequence of the residue theorem.  I will not prove it here, just state the result, which is if $f$ is sufficiently "well-behaved", then
$$\sum_{k=-\infty}^{\infty} (-1)^k \, f(k) = -\sum_n \operatorname*{Res}_{z=z_n} \left [\pi \, \csc{(\pi z)}\, f(z) \right ]$$
where the $z_n$ are non-integer poles of $f$ in the complex plane.
In our case,
$$f(z) = \frac1{2 z+1} \frac1{(2 z+1)^2+1} = \frac18 \frac1{z+\frac12} \frac1{\left (z+\frac12 \right )^2 + \frac14}$$
We have three poles: $z_1=-1/2$, $z_2 = -1/2 + i 1/2$, and $z_3=-1/2-i 1/2$.  Thus, the doubly infinite sum is equal to
$$-\pi \left [\csc{\left (-\frac{\pi}{2} \right )} \frac12 + \csc{\left (-\frac{\pi}{2} + i \frac{\pi}{2} \right )} \left (-\frac14 \right ) + \csc{\left (-\frac{\pi}{2} - i \frac{\pi}{2} \right )} \left (-\frac14 \right ) \right ] $$
which simplifies to
$$\frac{\pi}{2} \left [1-\operatorname{sech}{\left (\frac{\pi}{2} \right )} \right ] $$
Putting everything altogether, we have

$$\sum_{n=0}^{\infty} \sin{\left (n \frac{\pi}{2} \right )} \frac{n^2+2}{n^3+n} = \frac{\pi}{2} - \frac{\pi}{4}\, \operatorname{sech}{\left (\frac{\pi}{2} \right )}  $$

