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Does that sum converge absolutely, conditionally or diverges?

$$\sum_{n=2}^\infty\frac{1}{\ln^2{n}}\cos{\pi n^2}$$ I began with the absolute convergence and got to the point where I had to determine convergence of the sum: $$\sum_{n=2}^\infty\frac{1}{\ln^2{n}}$$ I thought about using comparison test, but it did not work.

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For sufficiently large $n$, $\ln{n} < \sqrt{n}$, so $(\ln{n})^2 < n$ and therefore $\frac{1}{(\ln{n})^2} > \frac{1}{n}$. $\sum_{n=2}^{\infty}\frac{1}{n}$ does not converge, so neither does $\sum_{n=2}^{\infty}\frac{1}{(\ln{n})^2}$.

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  • $\begingroup$ actually, not even $\sum_{n\geq 2} \frac{1}{nln\ n}$ converges $\endgroup$ – Veridian Dynamics Jan 30 '17 at 14:23
  • $\begingroup$ But $\sum_{n\ge2}\frac1{n\ln^2n}$ converges... $\endgroup$ – Simply Beautiful Art Jan 30 '17 at 14:29
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One can see with the Cauchy condensation test that

$$\sum_{n=1}^\infty\frac{2^n}{\ln^22^n}=\sum_{n=1}^\infty\frac{2^n}{n^2\ln^22}$$

which clearly diverges. Thus, if it converges, your series converges conditionally.

Noting that for integer $n$ that $\cos(\pi n^2)=(-1)^n$, it follows from the alternating test that the series converges.

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