91Ó°ÊÓ

To determine absolute convergence, consider the series \( \sum_{n=1}^{\infty} \left|(-1)^{n} \frac{(n !)^{2} 3^{n}}{(2 n+1) !}\right| = \sum_{n=1}^{\infty} \frac{(n !)^{2} 3^{n}}{(2 n+1) !} \). Use the Ratio Test to analyze this series:Show:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \]Here,\[ a_n = \frac{(n!)^2 3^n}{(2n+1)!} \]\[ a_{n+1} = \frac{((n+1)!)^2 3^{n+1}}{(2(n+1) + 1)!} = \frac{((n+1)!)^2 3^{n+1}}{(2n+3)!} \]Calculate the ratio:\[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{((n+1)!)^2 3^{n+1} \cdot (2n+1)!}{(n!)^2 3^n \cdot (2n+3)!} \]Simplify:\[ = \frac{((n+1)^2) 3 (2n+1)!}{(2n+3)(2n+2)(2n+1)!} \]\[ = \frac{3(n+1)^2}{(2n+3)(2n+2)} \]Find the limit:\[ \lim_{n \to \infty} \frac{3(n+1)^2}{(2n+3)(2n+2)} = \frac{3 \cdot \lim_{n \to \infty} (n+1)^2}{4n^2} = \frac{3}{4} \]Since \( \frac{3}{4} < 1 \), the series \( \sum_{n=1}^{\infty} \frac{(n !)^{2} 3^{n}}{(2 n+1) !} \) converges absolutely.

Since the series \( \sum_{n=1}^{\infty} \left| (-1)^{n} \frac{(n !)^{2} 3^{n}}{(2 n+1) !} \right| \) converges absolutely, the original series with the alternating sign, \( \sum_{n=1}^{\infty} (-1)^{n} \frac{(n !)^{2} 3^{n}}{(2 n+1) !} \), also converges. In general, any series that converges absolutely also converges.