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?

Without using asymptotic equivalence, an easy way is
\begin{equation}
\sum_{n=1}^\infty\left(\dfrac{n}{n+3}\times\dfrac{n+1}{n+3}\times\dfrac{1}{n+3}\right)\ge\sum_{n=1}^\infty\left(\dfrac14\times\dfrac24\times\dfrac{1}{n+3}\right)=\infty
\end{equation}
Another alternative solution is
\begin{equation}
\sum_{n=1}^\infty\dfrac{n(n+1)}{(n+3)^3}\ge\sum_{n=1}^\infty\dfrac{n^2}{(n+3n)^3}=\infty
\end{equation}
