<s>[INST] Converge or diverge $\sum^{\infty}_{n=2}\frac{4^n(n!)^2}{(2n-1)!}$ Show if the series converges or diverges.$$\sum^{\infty}_{n=2}\frac{4^n(n!)^2}{(2n-1)!}$$Can someone please help with proving this? (I think it converges)
 [/INST] $a_n=\dfrac{4^n(n!)^2}{(2n-1)!} = \dfrac{(2^n\cdot n!)^2}{(2n-1)!} = \dfrac{2\cdot 4\cdots (2n-2)\cdot (2n)}{1\cdot 2\cdot 3\cdots (2n-2)\cdot (2n-1)}\cdot 2\cdot 4\cdots (2n) > \dfrac{2\cdot 4\cdot 6\cdots (2n-2)\cdot (2n)}{1\cdot 2\cdot 3\cdots (2n-2)\cdot (2n-1)}\cdot 1\cdot 3\cdot 5\cdots (2n-1) = \dfrac{(2n-1)!2n}{(2n-1)!}= 2n = b_n \to \text{ divergence}$ by comparison test with $\sum_{n} (2n)$
</s>