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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?

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  • $\begingroup$ You can reduce the series. Show using Cauchy Criterion. $\endgroup$ – Melody Dec 3 '18 at 17:26
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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?
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  • $\begingroup$ The alternating series should converge right? $\endgroup$ – J. Lastin Dec 3 '18 at 18:05
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    $\begingroup$ @J.Lastin indeed $\endgroup$ – gt6989b Dec 3 '18 at 18:07
  • $\begingroup$ 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? $\endgroup$ – J. Lastin Dec 3 '18 at 18:09
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    $\begingroup$ @J.Lastin also correct $\endgroup$ – gt6989b Dec 3 '18 at 18:56
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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.

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