Convergence/divergence of series with sinus

Question

I want to know if this converges or diverges.

$\sin(n*\frac{\pi}{2})\frac{n^2 + 2}{n^3+n}$

I solved that $\frac{n^2 + 2}{n^3+n}$ diverges by limit comparasion test with $\frac{1}{n^2}$ but I dont know how to continue. Thanks for help.

Answer 1

The even $n$ terms are $0$, and can be discarded. The odd $n$ terms alternate in sign, since, for odd $n$, $\sin\left(\frac{\pi n}{2}\right)$ alternates between $1$ and $-1$.

Note that $\frac{n^2+2}{n^3+n}=\frac{n^2+1+1}{n(n^2+1)}=\frac{1}{n}+\frac{1}{n(n^2+1)}$. So the terms are decreasing in absolute value, and have limit $0$.

Thus by the Alternating Series Test, aka the Leibniz Test, our series converges.

With the work you did, we can say a little more. The series converges, but does not converge absolutely.