Problem 12. (6 points) Apply the Ratio Test to determine convergence or divergence, or state that the Ratio Test is inconclusive. \[ \sum_{n=1}^{\infty} \frac{7 n^{2}}{(2 n+1) !} \] $\rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\square($ Enter 'inf for $\infty$.) $\sum_{n=1}^{\infty} \frac{7 n^{2}}{(2 n+1) !}$ is: A. convergent B. divergent C. The Ratio Test is inconclusive Note: You can earn partial credit on this problem.

Step 1 ：Let \( a_n = \frac{7n^2}{(2n+1)!} \)

Step 2 ：Then \( a_{n+1} = \frac{7(n+1)^2}{(2(n+1)+1)!} = \frac{7(n+1)^2}{(2n+3)!} \)

Step 3 ：Calculate the ratio \( \frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{(2n+2)(2n+3)} \)

Step 4 ：Take the limit as \( n \) approaches infinity of the absolute value of the ratio, \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 \)

Step 5 ：Since the limit is less than 1, according to the Ratio Test, the series \( \sum_{n=1}^{\infty} \frac{7 n^{2}}{(2 n+1) !} \) is convergent

Step 6 ：Final Answer: \( \boxed{\text{A. convergent}} \)