confirming solution to the series $\frac{n(n+1)}{(n+3)^3}$ with ratio test I want to confirm my solution to this series using the ratio test correct, I tested to show the series is divergent.
$$\frac{n(n+1)}{(n+3)^3}$$
Using the ratio test, then simplifying in stages:
$$\frac{(n+1)(n+1+1)}{(n+1+3)^3}\frac{(n+3)^3}{n(n+1)}$$
$$\frac{(n+1)(n+2)(n+3)^3}{(n+4)^3n(n+1)}$$
$$\frac{(n+2)(n+3)^3}{(n+4)^2(n+4)n}$$
$$\frac{(n+3)^3}{(n+4)^22n}$$
The concluding remark:
$$\frac{1}{2}\lim_{n \to \infty}\frac{(n+3)^3}{(n+4)^2n}$$
Hence the series is divergent, unless I went wrong somewhere?

Ratio test is not effective because
$$\lim_{n\to \infty }\frac{(n+2) (n+3)^3}{n (n+4)^3}=1$$
You can use comparison test as
$$\frac{n (n+1)}{(n+3)^3}\sim \frac1n ;\text{ as }n\to\infty$$
and $\sum\frac1n $ diverges
