$\lim_{n\to \infty}\frac{(3^4)^{n+1}((n+1)!)^4}{(4(n+1))!}\cdot\frac{(4n)!}{(3^4)^n(n!)^4}$

Question

EDIT: This really is just a question about factorials, see @Angelo's answer

If it helps, I was originally trying to prove the convergence of $$\sum_{n=1}^{\infty}\frac{(3^4)^n(n!)^4}{(4n)!}$$ by using the ratio test...

I have to find:$$\lim_{n\to \infty}\frac{(3^4)^{n+1}((n+1)!)^4}{(4(n+1))!}\cdot\frac{(4n)!}{(3^4)^n(n!)^4}$$I'm pretty sure the issue comes down to how I simplified it:

$$\lim_{n\to \infty}\frac{(3^4)((n+1)!)^4}{(4(n+1))!}\cdot\frac{(4n)!}{(n!)^4}$$ $$\lim_{n\to \infty}\frac{(3^4)((n+1)n!)^4}{(4n+4)!}\cdot\frac{(4n)!}{(n!)^4}$$ $$\lim_{n\to \infty}\frac{(3^4)(n+1)^4}{(4n+4)(4n)!}\cdot(4n)!$$ $$\lim_{n\to \infty}\frac{3^4}{4}\cdot(n+1)^3$$=$\infty$, so the series diverges. But according to my homework, that's incorrect.

Any help is appreciated

Answer 1

Actually

$$\lim_{n\to \infty}\frac{(3^4)\big((n+1)n!\big)^4}{(4n+4)!}\cdot\frac{(4n)!}{(n!)^4}=$$

$$=\lim_{n\to \infty}\frac{(3^4)(n+1)^4}{(4n+4)(4n+3)(4n+2)(4n+1)(4n)!}\cdot(4n)!=$$

$$=\lim_{n\to \infty}\frac{3^4(n+1)^4}{4^4(n+1)\left(n+\frac34\right)\left(n+\frac12\right)\left(n+\frac14\right)}=\left(\frac34\right)^4<1\;.$$

So the series converges.

Answer 2

You made a mistake in your calculation. It should be:$$(4n+4)!=(4n+4)(4n+3)(4n+2)(4n+1)(4n!)$$