Answer
converges
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{1}{n 2^{n}+n^{3}}$$
Since $a_n = \dfrac{1}{n 2^{n}+n^{3}}$, $a_{n+1} = \dfrac{1}{(n+1) 2^{n+1}+(n+1)^{3}}$, then by using the Ratio test
\begin{align*}
\rho &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\
&=\lim _{n \rightarrow \infty}\frac{n 2^{n}+n^{3}}{(n+1) 2^{n+1}+(n+1)^{3}}\\
&=\lim _{n \rightarrow \infty}\frac{n 2^{n}\left(1+\frac{n^{2}}{2^{n}}\right)}{(n+1) 2^{n+1}\left(1+\frac{(n+1)^{2}}{2^{n+1}}\right)}\\
&=\frac{1}{2} \lim _{n \rightarrow \infty} \frac{n}{n+1} \lim _{n \rightarrow \infty}\frac{1+\frac{n^{2}}{2^{n}}}{1+\frac{(n+1)^{2}}{2^{n+1}}}\\
&=\frac{1}{2}(1)(1)=\frac{1}{2}<1
\end{align*}
Then the given series converges.
