Question

Compute the series

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\binom{2n}{2}} = \frac{1}{\binom{2}{2}} - \frac{1}{\binom{4}{2}} + \frac{1}{\binom{6}{2}} - \frac{1}{\binom{8}{2}} + \frac{1}{\binom{10}{2}} + \frac{1}{\binom{12}{2}} + \dots$

Answer 1

$\frac{\pi}{2} - \ln(2)$

Answer 2

The series term $\frac{(-1)^{n-1}}{\binom{2n}{2}}$ is simplified using $\binom{2n}{2} = n(2n-1)$. The resulting fraction $\frac{1}{n(2n-1)}$ is decomposed into partial fractions $\frac{2}{2n-1} - \frac{1}{n}$. This transforms the original series into two known alternating series: $2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n-1}$ and $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$. The first sum is $2 \times \arctan(1) = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$. The second sum is $\ln(1+1) = \ln(2)$. Combining these results yields $\frac{\pi}{2} - \ln(2)$.