<s>[INST] Simplified expression of the ratio of this series I need to simplify the expression of this series:
$$
\sum\limits_{n=1}^{\infty}\frac{3^{n}(4n)!}{1 · 4 · 7 · ⋯ ·3n+1}
$$
As the ratio (for the test): $$\frac{a_{n+1}}{a_n}$$
First I multiplied the reciprocal of a(n) by a(n+1):
$$
\frac{a_{n+1}}{a_n}=\frac{3^{n+1}(4(n+1))!}{(3(n+1)+1)}\cdot\frac{3n+1}{3^{n}(4n)!}
$$
Then I expanded the terms:
$$
\frac{3^{n+1}\cdot4n!(4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{3^{n}\cdot4n!(3n+4)(3n+3)(3n+2)(3n+1)}
$$
Simplified down to:
$$
\frac{a_{n+1}}{a_n}=\frac{3(4n+4)(4n+3)(4n+2)(4n+1)}{(3n+4)(3n+3)(3n+2)}
$$
Am I missing a step or performing any incorrectly?
This question is from a practice test, which is multiple choice, but I don't have an answer key:
See choices here
 [/INST] \begin{align}
\frac{a_{n+1}}{a_n}&=
\frac{3^{n+1}\cdot4n!(4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{3^{n}\cdot 4n!(3n+4)}\\
&=
\frac{3 (4n+4)(4n+3)(4n+2)(4n+1)(3n+1)}{(3n+4)}.
\end{align}
Answer F.
</s>