How to choose $b_n$ for $\sum_{n=1}^{\infty}\frac{5n^3+1}{2^n\left(n^3+n+1\right)}$ So I have to use the limit comparison test for the following problem, but I'm struggling with how to choose my $b_n$.
$$\sum_{n=1}^{\infty}\frac{5n^3+1}{2^n\left(n^3+n+1\right)}$$
My first thought was to compare $2^n$ and $(n^3+n+1)$ to see which is faster growing, but then I wasn't sure, since these were in the denominator if I should actually compare $\frac{1}{2^n}$ and $\frac{1}{n^3+n+1}$, to see which goes to zero the slowest.
But, then I realize that all the examples I had for choosing terms in this fashion were additive, and not multiplicative.  So I'm really not sure what the best way to choose $b_n$ is for this problem.  Any tips?
Thank you!

So I think I figured this out on my own, just wanted to post and see if this looks right:
$$\sum_{n=1}^{\infty}\frac{5n^3+1}{2^n\left(n^3+n+1\right)}\xrightarrow[]{dominant\ power\ terms}b_n=\frac{5n^3}{2^nn^3}=5\left ( \frac{1}{2} \right )^n\rightarrow Gradient\ series\ where \left | r \right |< 1\rightarrow convergent$$
then by LCT:
$$\lim_{n \to \infty}\frac{\frac{5n^3+1}{2^n(n^3+n+1)}}{\frac{5}{2^n}}\rightarrow \lim_{n \to \infty}\frac{5n^3+1}{2^n(n^3+n+1)}\ast \frac{2^n}{5}\rightarrow \lim_{n \to \infty}\frac{n^3+\frac{1}{5}}{n^3+n+1}\xrightarrow[]{dominant\ terms}\lim_{n \to \infty}\frac{n^3}{n^3}=1\rightarrow both\ series\ converge$$
