# Convergence of the following sum.

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.

## 2 Answers

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 $\sum_{n=2}^{\infty}\frac{1}{(\ln{n})^2}$.

• actually, not even $\sum_{n\geq 2} \frac{1}{nln\ n}$ converges – Veridian Dynamics Jan 30 '17 at 14:23
• But $\sum_{n\ge2}\frac1{n\ln^2n}$ converges... – Simply Beautiful Art Jan 30 '17 at 14:29

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.