## Calculus (3rd Edition)

The series $\sum_{n=2}^{\infty} \frac{\cos n \pi}{(\ln n)^{2}}$ converges conditionally.
Since $\cos (n\ pi)$ is either $1$ or $-1$ (depending on whether $n$ is even or odd), then we can write the series $\sum_{n=2}^{\infty} \frac{\cos n \pi}{(\ln n)^{2}}$ in the form $\sum_{n=2}^{\infty} \frac{(-1)^n}{(\ln n)^{2}}$. Now, we study the positive series $\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}}$. Using the property that $\ln n \lt n^a, \ \ a\gt 0$, then we have $\ln n \lt n^{1/2}$ then $(\ln n)^{2}\lt n$, now we have $$\frac{1}{(\ln n)^{2}}\geq\frac{ 1}{n}$$ Since $\Sigma_{n=2}^{\infty}\frac{ 1}{n}$ is a divergent p-series ( $p=1$) then the series $\Sigma_{n=2}^{\infty} \frac{1}{(\ln n)^{2}}$ diverges. Also, $$\lim_{n\to\infty}b_n=\lim_{n\to\infty} \frac{1}{(\ln n)^{2}}=0.$$ So, by using the alternating series test, the series $\sum_{n=2}^{\infty} \frac{\cos n \pi}{(\ln n)^{2}}$ converges conditionally.