For the following Series, decide whether they converge or diverge. \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2 n-1} \) \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{(n!)^{2}} \) \( \sum_{n=1}^{\alpha}(-1)^{(n+1)} \frac{1}{(2 n-1)^{2}} \)

To determine whether the given series converge or diverge, we can apply the Alternating Series Test and other convergence tests as appropriate. 1. **Series**: \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2 n-1} \) This is an alternating series of the form \( \sum (-1)^{n+1} a_n \) where \( a_n = \frac{1}{2n-1} \). To apply the Alternating Series Test, we need to check two conditions: - \( a_n \) is positive: \( a_n = \frac{1}{2n-1} > 0 \) for all \( n \geq 1 \). - \( a_n \) is decreasing: We check if \( a_{n+1} < a_n \): \[ a_{n+1} = \frac{1}{2(n+1)-1} = \frac{1}{2n+1} \] Since \( 2n+1 > 2n-1 \), we have \( a_{n+1} < a_n \). - \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{2n-1} = 0 \). Since both conditions are satisfied, the series converges by the Alternating Series Test. 2. **Series**: \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{(n!)^{2}} \) This is also an alternating series where \( a_n = \frac{1}{(n!)^2} \). Again, we check the conditions for the Alternating Series Test: - \( a_n \) is positive: \( a_n = \frac{1}{(n!)^2} > 0 \) for all \( n \geq 1 \). - \( a_n \) is decreasing: We need to show \( a_{n+1} < a_n \): \[ a_{n+1} = \frac{1}{((n+1)!)^2} = \frac{1}{((n+1) \cdot n!)^2} = \frac{1}{(n+1)^2 (n!)^2} \] Since \( (n+1)^2 > 1 \), we have \( a_{n+1} < a_n \). - \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{(n!)^2} = 0 \). Since both conditions are satisfied, the series converges by the Alternating Series Test. 3. **Series**: \( \sum_{n=1}^{\alpha}(-1)^{(n+1)} \frac{1}{(2 n-1)^{2}} \) This series is finite because it sums from \( n=1 \) to \( n=\alpha \), where \( \alpha \) is a finite number. Any finite sum converges, regardless of whether the terms are positive or negative. In summary: 1. \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2 n-1} \) converges. 2. \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{(n!)^{2}} \) converges. 3. \( \sum_{n=1}^{\alpha}(-1)^{(n+1)} \frac{1}{(2 n-1)^{2}} \) converges (as it is a finite sum).