<s>[INST] 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?
 [/INST] 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

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

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