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How can I figure out if this series converges absolutely?

$$ \sum_{n=1}^\infty \frac {\cos(n \pi)}{(n+1)\ln(n+1)} $$

The ratio and root test are both inconclusive (according to Wolfram Alpha).

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    $\begingroup$ How much is $\cos(n\pi)$, already? $\endgroup$ – Did Jun 14 '17 at 16:15
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hint

$$\cos (n\pi)=(-1)^n $$

using alternate series test , it converges but using comparison with integral, it is not absolutely convergent :

$$\lim_{X\to+\infty}\int_1^X\frac {dt}{(t+1)\ln (t+1)}=$$

$$\lim_{X\to+\infty}\Big [\ln (\ln (t+1))\Bigr]_1^X =+\infty$$

thus, the series $$\sum \frac {1}{(n+1)\ln (n+1)} $$ is divergent.

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  • $\begingroup$ Absolute convergence. $\endgroup$ – Dando18 Jun 14 '17 at 16:22
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Hints:

1) $\cos n\pi = (-1)^n \quad\forall n\in \mathbb{N}$

2) $\lvert (-1)^n \rvert = 1$

3) $\frac{1}{(n+1)\log(n+1)} \sim \frac{1}{n\log n}$ (limit comparison test)

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  • $\begingroup$ We cannot conclude by comparison test. $\endgroup$ – hamam_Abdallah Jun 15 '17 at 1:02

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