Series convergence: $sin (n \frac{\pi}{2})$

Determine whether the following series : $$\sum_{n=1}^\infty \sin \left(\frac{n\pi}{2}\right) \frac{n^2+2}{n^3 +n}$$ converges absolutely, conditionally or diverges.

I know that for even natural numbers the expression will equal zero and that for odd values of $$n$$ the value of $$\sin$$ will go from $$1$$ to $$-1$$.

Could I theoretically reduce this series into a subseries:

$$\sum_{n=0}^\infty \sin \left(\frac{(2n+1)\pi}{2}\right) \frac{(2n+1)^2+2}{(2n+1)^3 +2n + 1}$$

And then treat it as if it were a standard alternating series?

• You can reduce the series. Show using Cauchy Criterion. – Melody Dec 3 '18 at 17:26

HINT

• Note that $$\frac{n^2+2}{n^3+n} = \frac{1}{n} \times \frac{n^2+2}{n^2+1} = \frac{1}{n} \left[ 1 + \frac{1}{n^2+1} \right] = \Theta(1/n)$$ Does this series converge?
• Adding $$\sin(n\pi/2)$$ in the front effectively kills all the even-$$n$$ terms and makes an alternating series out of the odd ones -- does the alternating series converge?
• The alternating series should converge right? – J. Lastin Dec 3 '18 at 18:05
• @J.Lastin indeed – gt6989b Dec 3 '18 at 18:07
• Okay and if i want to check for absolute convergence, can I do limit comparison with $\frac{1}{n}$, meaning that the series converges only conditionally? – J. Lastin Dec 3 '18 at 18:09
• @J.Lastin also correct – gt6989b Dec 3 '18 at 18:56

Guide:

• We have $$\sum_{n=1}^\infty \sin (\frac{n\pi}{2}) \frac{n^2+2}{n^3 +n}= \sum_{n=1}^\infty(-1)^{n+1} \frac{(2n-1)^2+2}{(2n-1)^3 +(2n-1)}$$

• Try alternating series test.