## Calculus (3rd Edition)

The series $\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{1 / 2}(\ln n)^{2}}$ converges.
For the series $\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{1 / 2}(\ln n)^{2}}$, the positive term $b_n=\frac{1}{n^{1 / 2}(\ln n)^{2}}$, which is positive and decreasing. Evaluate the limit: $$\lim_{n\to \infty}b_n=\lim_{n\to \infty}\frac{1}{n^{1 / 2}(\ln n)^{2}}=\frac{1}{\infty}=0$$ Hence, using the alternating series test, the series $\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{1 / 2}(\ln n)^{2}}$ converges.