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$\sum\frac{(-1)^{n+1}}{2n^2-3n\cos(n\pi)}$ show that this is convergence or divergence.

Ok i couldnt find a way to show this using the alternating series test. And when i used the ratio test i found that it was equal or smaller than one meaning i cant draw a conclusion. ratio test:

lim$\frac{2n^2-3n\cos(n\pi)}{2(n+1)^2-3(n+1)\cos(n\pi)}$ this is equal or smaller than lim$\frac{2n^2+3n}{2n^2-3n}=1$ so it is smaller or equal than 1 so it doesnt help me.

What should i do somebody can help me?

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  • $\begingroup$ @ Danielvanheuven Your sum has a closed expression $\frac{1}{6} (-\pi +4-4 \log (2)) \simeq -0.31903$ $\endgroup$ Dec 6, 2019 at 9:58

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For $n \geq 3$ we have $|\frac {(-1)^{n+1}} {2n^{2}-3n\cos (n\pi)}|\leq \frac 1 {2n^{2}-3n} \leq \frac 1 {n^{2}}$. Hence the series is absolutely convergent.

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    $\begingroup$ @dmtri That was a typo. Thanks for pointing out. $\endgroup$ Dec 6, 2019 at 9:47
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Note that$$n>1\implies\left\lvert\frac{(-1)^{n+1}}{2n^2-3n\cos(n\pi)}\right\rvert\leqslant\frac1{2n^2-3n}$$and that the series $\sum_{n=2}^\infty\frac1{2n^2-3n}$ converges.

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