ratio test to find series convergence

Question

We have a series

$$b_n=\sum_{n=1}^{\infty} \frac{(3n^3+1)^36^n}{3(n!)}$$

I assume to use positive ratio test $\frac{b_{n+1}}{b_n}$ to find if it's convergent.

$$ \begin{align} \frac{b_{n+1}}{b_n}&=\frac{(3n^3+2)^36^{n+1}}{3(n+1)!} \cdot \frac{3(n!)}{(3n^3+1)^36^n}\\ &=\frac{6(3n^3+2)^3}{(3n^3+1)^3(n+1)} \end{align} $$

What method do I then use to then to prove that it's sequence converges to limit 0?

Answer 1

$$\lim_{n \to \infty} \frac{b_{n+1}}{b_n} =\frac{6(3n^3+2)^3}{(3n^3+1)^3(n+1)} = \frac{6 (3+\frac 2n)^3}{(3+\frac 1n )^3 (n+1)} \to \frac{6 \cdot 27}{27(n+1)} =0$$

Hence, series converges.

Answer 2

one factor is given by $$\frac{n^9\left(3+\frac{2}{n^3}\right)^3}{n^9\left(3+\frac{1}{n^3}\right)^3}$$

Answer 3

Note that the polynomial term in negligeble since by ratio test it gives a constant term indeed

$$\frac{ (3(n+1)^3+1)^3 }{(3n^3+1)^3}\to 1$$

thus you can consider the leading term $c_n=\frac{6^n}{n!}$

$$\frac{c_{n+1}}{c_n}=\frac{6^{n+1}}{(n+1)!}\frac{n!}{6^n}=\frac{6}{n+1}\to 0$$

thus $\frac{b_{n+1}}{b_n}\to0$ and the given series converges.